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with respect to $x$,","difficulty":1,"difficultyBand":"easy","tier":null,"totalMarks":4,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"ebbdc335-2184-49ea-a5c0-97be5df31dea","label":"a","content":"$$6x^2 + e^{2x}$","markscheme":"- Differentiates $6x^2$ to $12x$ B1\n- Differentiates $e^{2x}$ to $2e^{2x}$ B1","marks":2,"order":0,"markingRubric":{"totalMarks":2,"points":[{"id":"p1","marks":1,"description":"Differentiates $6x^2$ correctly to $12x$","code":"B"},{"id":"p2","marks":1,"description":"Differentiates $e^{2x}$ correctly to $2e^{2x}$","code":"B"}],"finalAnswers":["12x + 2e^{2x}"],"generatedAt":"2026-05-06T14:03:50.808Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"direct_differentiation","label":"Direct Differentiation","steps":[{"type":"text","order":0,"title":"Apply the sum rule","content":"To differentiate the expression $6x^2 + e^{2x}$, we can use the sum rule of differentiation. This rule states that the derivative of a sum is the sum of the derivatives:\n\n$$\\frac{d}{dx}\\left(6x^2 + e^{2x}\\right) = \\frac{d}{dx}\\left(6x^2\\right) + \\frac{d}{dx}\\left(e^{2x}\\right)$$\n\nWe will differentiate each term individually."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\frac{d}{dx}\\left(\\htmlData{anchor=alg-term}{6x^2}\\right) = 6 \\cdot \\htmlData{anchor=power-rule}{2x^{2-1}}$"},{"latex":"$$\\phantom{\\frac{d}{dx}\\left(6x^2\\right)} = \\htmlData{anchor=alg-result}{12x}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"alg-term"},{"color":"green","style":"box","anchor":"alg-result"}],"connections":[{"to":"power-rule","from":"alg-term","color":"sky","label":"power rule","style":"solid"}]},"order":1,"title":"Differentiate the algebraic term","description":"First, we differentiate $6x^2$ using the power rule, $\\frac{d}{dx}(x^n) = nx^{n-1}$. This earns you your first mark **(B1)**."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\frac{d}{dx}\\left(\\htmlData{anchor=exp-term}{e^{2x}}\\right) = \\htmlData{anchor=chain-rule}{2} \\cdot e^{2x}$"},{"latex":"$$\\phantom{\\frac{d}{dx}\\left(e^{2x}\\right)} = \\htmlData{anchor=exp-result}{2e^{2x}}$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"exp-term"},{"color":"green","style":"box","anchor":"exp-result"}],"connections":[{"to":"chain-rule","from":"exp-term","color":"amber","label":"chain rule","style":"solid"}]},"order":2,"title":"Differentiate the exponential term","description":"Next, we differentiate $e^{2x}$ using the chain rule, $\\frac{d}{dx}\\left(e^{f(x)}\\right) = f'(x)e^{f(x)}$. Since the derivative of $2x$ is $2$, we multiply the exponential function by $2$. This earns you your second mark **(B1)**."},{"type":"text","order":3,"title":"Combine the results","content":"Combining both differentiated terms together, we get the final derivative:\n\n$$12x + 2e^{2x}$$"}]}],"generatedAt":"2026-06-09T01:56:49.265Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"e6811698-1c26-47c5-82b0-bf8f153fe15d","label":"b","content":"$$(x^2 - 3)^5$","markscheme":"- Applies chain rule: $5(x^2 - 3)^4 \\times \\dots$ M1\n- Correct derivative $10x(x^2 - 3)^4$ A1","marks":2,"order":1,"markingRubric":{"totalMarks":2,"points":[{"id":"p1","marks":1,"description":"Applies chain rule leading to $5(x^2 - 3)^4 \\dots$","code":"M"},{"id":"p2","marks":1,"description":"Completes derivative to obtain $10x(x^2 - 3)^4$","code":"A"}],"finalAnswers":["10x(x^2 - 3)^4"],"generatedAt":"2026-05-06T14:03:50.808Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"direct_chain_rule","label":"Direct Chain Rule","steps":[{"type":"text","order":0,"title":"Identify the chain rule","content":"To differentiate the function $(x^2 - 3)^5$, we use the **Direct Chain Rule**.\n\nThis method requires us to:\n1. Apply the power rule to the **outer function**, $(\\text{inner})^5$, keeping the inner function unchanged.\n2. Multiply directly by the derivative of the **inner function**, $x^2 - 3$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y = (\\htmlData{anchor=inner}{\\color{@sky}{x^2 - 3}})^{\\htmlData{anchor=power}{\\color{@purple}{5}}}$"},{"latex":"$$\\frac{dy}{dx} = \\htmlData{anchor=outer-deriv}{\\color{@purple}{5}(\\color{@sky}{x^2 - 3})^{\\color{@purple}{4}}} \\times \\frac{d}{dx}(\\htmlData{anchor=inner-to-deriv}{\\color{@sky}{x^2 - 3}})$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"inner"},{"color":"purple","style":"text-color","anchor":"power"},{"color":"purple","style":"text-color","anchor":"outer-deriv"},{"color":"sky","style":"text-color","anchor":"inner-to-deriv"}],"connections":[{"to":"outer-deriv","from":"power","color":"purple","label":"power rule","style":"solid"},{"to":"inner-to-deriv","from":"inner","color":"sky","label":"inner part","style":"solid"}]},"order":1,"title":"Differentiate outer function","description":"First, we apply the power rule to the outer power of $5$, leaving the inside expression $x^2 - 3$ exactly as it is. This gives us the first part of our derivative, which earns the first mark."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\frac{d}{dx}(\\htmlData{anchor=inner-term}{\\color{@sky}{x^2 - 3}}) = \\htmlData{anchor=inner-deriv}{\\color{@amber}{2x}}$"},{"latex":"$$\\frac{dy}{dx} = 5(x^2 - 3)^4 \\times \\htmlData{anchor=inner-sub}{\\color{@amber}{2x}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"inner-term"},{"color":"amber","style":"text-color","anchor":"inner-deriv"},{"color":"amber","style":"text-color","anchor":"inner-sub"}],"connections":[{"to":"inner-sub","from":"inner-deriv","color":"amber","label":"substitute","style":"solid"}]},"order":2,"title":"Multiply by inner derivative","description":"Now, we find the derivative of the inner function, $x^2 - 3$, which is $2x$. We multiply our previous result by this derivative."},{"type":"text","order":3,"title":"Simplify final answer","content":"Finally, we simplify the expression by multiplying the coefficients together:\n\n$$\\frac{dy}{dx} = 5 \\times 2x(x^2 - 3)^4 = 10x(x^2 - 3)^4$$\n\nThis completes our differentiation and secures the final mark."}]},{"id":"chain_rule_substitution","label":"Chain Rule with Substitution","steps":[{"type":"text","order":0,"title":"Introduce the substitution","content":"To differentiate $y = (x^2 - 3)^5$ using the chain rule, we can introduce a new variable $u$ for the inner function.\n\nLet:\n$$u = x^2 - 3$$\n\nThis allows us to write $y$ as a simpler function of $u$:\n$$y = u^5$$"},{"type":"text","order":1,"title":"Differentiate each part","content":"Now, we find the derivatives of $u$ with respect to $x$, and $y$ with respect to $u$:\n\n1. Differentiating $u = x^2 - 3$ with respect to $x$ gives:\n$$\\frac{du}{dx} = 2x$$\n\n2. Differentiating $y = u^5$ with respect to $u$ using the power rule gives:\n$$\\frac{dy}{du} = 5u^4$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=dydu}{\\color{@sky}{\\frac{dy}{du}}} = \\color{@sky}{5u^4} \\quad \\text{and} \\quad \\htmlData{anchor=dudx}{\\color{@amber}{\\frac{du}{dx}}} = \\color{@amber}{2x}$"},{"latex":"$$\\frac{dy}{dx} = \\htmlData{anchor=formula-dydu}{\\color{@sky}{\\frac{dy}{du}}} \\times \\htmlData{anchor=formula-dudx}{\\color{@amber}{\\frac{du}{dx}}}$"},{"latex":"$$\\frac{dy}{dx} = (\\color{@sky}{5u^4}) \\times (\\color{@amber}{2x})$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"dydu"},{"color":"amber","style":"text-color","anchor":"dudx"}],"connections":[{"to":"formula-dydu","from":"dydu","color":"sky","label":"substitute","style":"solid"},{"to":"formula-dudx","from":"dudx","color":"amber","label":"substitute","style":"solid"}]},"order":2,"title":"Apply the chain rule","description":"We use the chain rule formula to combine our derivatives: $\\frac{dy}{dx} = \\frac{dy}{du} \\times \\frac{du}{dx}$."},{"type":"text","order":3,"title":"Substitute back and simplify","content":"To finish, we substitute $u = x^2 - 3$ back into our expression and simplify:\n\n$$\\begin{aligned} \\frac{dy}{dx} &= 5(x^2 - 3)^4 \\times 2x \\\\ &= 10x(x^2 - 3)^4 \\end{aligned}$$\n\nThis matches the expected derivative of $10x(x^2 - 3)^4$ in the markscheme, earning the method mark (M1) for applying the chain rule, and the accuracy mark (A1) for the correct final simplified expression."}]},{"id":"binomial_expansion","label":"Binomial Expansion and Term-by-Term Differentiation","steps":[{"type":"text","order":0,"title":"State the method","content":"We need to differentiate $(x^2 - 3)^5$ with respect to $x$.\n\nWhile this derivative is commonly found using the chain rule, we will solve it using **binomial expansion** to fully expand the expression first, followed by **term-by-term differentiation**."},{"type":"text","order":1,"title":"Apply Binomial Theorem","content":"Recall the binomial expansion formula:\n\n$$(a + b)^n = \\sum_{k=0}^{n} \\binom{n}{k} a^{n-k} b^k$$\n\nHere, we have $a = x^2$, $b = -3$, and $n = 5$. Let's write out the terms:\n\n$$\\begin{aligned} (x^2 - 3)^5 = \\;& \\binom{5}{0}(x^2)^5(-3)^0 + \\binom{5}{1}(x^2)^4(-3)^1 + \\binom{5}{2}(x^2)^3(-3)^2 \\\\ &+ \\binom{5}{3}(x^2)^2(-3)^3 + \\binom{5}{4}(x^2)^1(-3)^4 + \\binom{5}{5}(x^2)^0(-3)^5 \\end{aligned}$$"},{"type":"text","order":2,"title":"Simplify the expansion","content":"Now we calculate the coefficients and simplify each term using the binomial coefficients:\n* $\\binom{5}{0} = 1$, $\\binom{5}{1} = 5$, $\\binom{5}{2} = 10$, $\\binom{5}{3} = 10$, $\\binom{5}{4} = 5$, $\\binom{5}{5} = 1$\n\nSubstituting these values in, we get:\n\n$$\\begin{aligned} (x^2 - 3)^5 &= 1(x^{10})(1) + 5(x^8)(-3) + 10(x^6)(9) + 10(x^4)(-27) + 5(x^2)(81) + 1(1)(-243) \\\\ &= x^{10} - 15x^8 + 90x^6 - 270x^4 + 405x^2 - 243 \\end{aligned}$$"},{"type":"text","order":3,"title":"Differentiate term-by-term","content":"Now we apply the power rule for differentiation, $\\frac{d}{dx}(x^n) = nx^{n-1}$, to each term of the expanded polynomial:\n\n$$\\begin{aligned} \\frac{d}{dx}\\left[(x^2 - 3)^5\\right] &= \\frac{d}{dx}\\left(x^{10} - 15x^8 + 90x^6 - 270x^4 + 405x^2 - 243\\right) \\\\ &= 10x^9 - 15(8x^7) + 90(6x^5) - 270(4x^3) + 405(2x) - 0 \\\\ &= 10x^9 - 120x^7 + 540x^5 - 1080x^3 + 810x \\end{aligned}$$"},{"type":"text","order":4,"title":"Factorise the derivative","content":"To check our result against the standard chain rule form, let's factor out the common factor $10x$:\n\n$$\\frac{d}{dx}\\left[(x^2 - 3)^5\\right] = 10x\\left(x^8 - 12x^6 + 54x^4 - 108x^2 + 81\\right)$$\n\nNotice that the term in the brackets is the perfect fourth-power expansion of $(x^2 - 3)^4$:\n\n$$(x^2 - 3)^4 = x^8 - 12x^6 + 54x^4 - 108x^2 + 81$$\n\nSubstituting this back gives our final simplified answer:\n\n$$10x(x^2 - 3)^4$$"}]}],"generatedAt":"2026-06-09T02:01:11.181Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"ebbdc335-2184-49ea-a5c0-97be5df31dea:direct_differentiation","label":"(a) Direct Differentiation","steps":[{"type":"text","order":0,"title":"Apply the sum rule","content":"To differentiate the expression $6x^2 + e^{2x}$, we can use the sum rule of differentiation. This rule states that the derivative of a sum is the sum of the derivatives:\n\n$$\\frac{d}{dx}\\left(6x^2 + e^{2x}\\right) = \\frac{d}{dx}\\left(6x^2\\right) + \\frac{d}{dx}\\left(e^{2x}\\right)$$\n\nWe will differentiate each term individually."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\frac{d}{dx}\\left(\\htmlData{anchor=alg-term}{6x^2}\\right) = 6 \\cdot \\htmlData{anchor=power-rule}{2x^{2-1}}$"},{"latex":"$$\\phantom{\\frac{d}{dx}\\left(6x^2\\right)} = \\htmlData{anchor=alg-result}{12x}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"alg-term"},{"color":"green","style":"box","anchor":"alg-result"}],"connections":[{"to":"power-rule","from":"alg-term","color":"sky","label":"power rule","style":"solid"}]},"order":1,"title":"Differentiate the algebraic term","description":"First, we differentiate $6x^2$ using the power rule, $\\frac{d}{dx}(x^n) = nx^{n-1}$. This earns you your first mark **(B1)**."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\frac{d}{dx}\\left(\\htmlData{anchor=exp-term}{e^{2x}}\\right) = \\htmlData{anchor=chain-rule}{2} \\cdot e^{2x}$"},{"latex":"$$\\phantom{\\frac{d}{dx}\\left(e^{2x}\\right)} = \\htmlData{anchor=exp-result}{2e^{2x}}$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"exp-term"},{"color":"green","style":"box","anchor":"exp-result"}],"connections":[{"to":"chain-rule","from":"exp-term","color":"amber","label":"chain rule","style":"solid"}]},"order":2,"title":"Differentiate the exponential term","description":"Next, we differentiate $e^{2x}$ using the chain rule, $\\frac{d}{dx}\\left(e^{f(x)}\\right) = f'(x)e^{f(x)}$. Since the derivative of $2x$ is $2$, we multiply the exponential function by $2$. This earns you your second mark **(B1)**."},{"type":"text","order":3,"title":"Combine the results","content":"Combining both differentiated terms together, we get the final derivative:\n\n$$12x + 2e^{2x}$$"}]},{"id":"e6811698-1c26-47c5-82b0-bf8f153fe15d:direct_chain_rule","label":"(b) Direct Chain Rule","steps":[{"type":"text","order":0,"title":"Identify the chain rule","content":"To differentiate the function $(x^2 - 3)^5$, we use the **Direct Chain Rule**.\n\nThis method requires us to:\n1. Apply the power rule to the **outer function**, $(\\text{inner})^5$, keeping the inner function unchanged.\n2. Multiply directly by the derivative of the **inner function**, $x^2 - 3$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y = (\\htmlData{anchor=inner}{\\color{@sky}{x^2 - 3}})^{\\htmlData{anchor=power}{\\color{@purple}{5}}}$"},{"latex":"$$\\frac{dy}{dx} = \\htmlData{anchor=outer-deriv}{\\color{@purple}{5}(\\color{@sky}{x^2 - 3})^{\\color{@purple}{4}}} \\times \\frac{d}{dx}(\\htmlData{anchor=inner-to-deriv}{\\color{@sky}{x^2 - 3}})$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"inner"},{"color":"purple","style":"text-color","anchor":"power"},{"color":"purple","style":"text-color","anchor":"outer-deriv"},{"color":"sky","style":"text-color","anchor":"inner-to-deriv"}],"connections":[{"to":"outer-deriv","from":"power","color":"purple","label":"power rule","style":"solid"},{"to":"inner-to-deriv","from":"inner","color":"sky","label":"inner part","style":"solid"}]},"order":1,"title":"Differentiate outer function","description":"First, we apply the power rule to the outer power of $5$, leaving the inside expression $x^2 - 3$ exactly as it is. This gives us the first part of our derivative, which earns the first mark."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\frac{d}{dx}(\\htmlData{anchor=inner-term}{\\color{@sky}{x^2 - 3}}) = \\htmlData{anchor=inner-deriv}{\\color{@amber}{2x}}$"},{"latex":"$$\\frac{dy}{dx} = 5(x^2 - 3)^4 \\times \\htmlData{anchor=inner-sub}{\\color{@amber}{2x}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"inner-term"},{"color":"amber","style":"text-color","anchor":"inner-deriv"},{"color":"amber","style":"text-color","anchor":"inner-sub"}],"connections":[{"to":"inner-sub","from":"inner-deriv","color":"amber","label":"substitute","style":"solid"}]},"order":2,"title":"Multiply by inner derivative","description":"Now, we find the derivative of the inner function, $x^2 - 3$, which is $2x$. We multiply our previous result by this derivative."},{"type":"text","order":3,"title":"Simplify final answer","content":"Finally, we simplify the expression by multiplying the coefficients together:\n\n$$\\frac{dy}{dx} = 5 \\times 2x(x^2 - 3)^4 = 10x(x^2 - 3)^4$$\n\nThis completes our differentiation and secures the final mark."}]},{"id":"e6811698-1c26-47c5-82b0-bf8f153fe15d:chain_rule_substitution","label":"(b) Chain Rule with Substitution","steps":[{"type":"text","order":0,"title":"Introduce the substitution","content":"To differentiate $y = (x^2 - 3)^5$ using the chain rule, we can introduce a new variable $u$ for the inner function.\n\nLet:\n$$u = x^2 - 3$$\n\nThis allows us to write $y$ as a simpler function of $u$:\n$$y = u^5$$"},{"type":"text","order":1,"title":"Differentiate each part","content":"Now, we find the derivatives of $u$ with respect to $x$, and $y$ with respect to $u$:\n\n1. Differentiating $u = x^2 - 3$ with respect to $x$ gives:\n$$\\frac{du}{dx} = 2x$$\n\n2. Differentiating $y = u^5$ with respect to $u$ using the power rule gives:\n$$\\frac{dy}{du} = 5u^4$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=dydu}{\\color{@sky}{\\frac{dy}{du}}} = \\color{@sky}{5u^4} \\quad \\text{and} \\quad \\htmlData{anchor=dudx}{\\color{@amber}{\\frac{du}{dx}}} = \\color{@amber}{2x}$"},{"latex":"$$\\frac{dy}{dx} = \\htmlData{anchor=formula-dydu}{\\color{@sky}{\\frac{dy}{du}}} \\times \\htmlData{anchor=formula-dudx}{\\color{@amber}{\\frac{du}{dx}}}$"},{"latex":"$$\\frac{dy}{dx} = (\\color{@sky}{5u^4}) \\times (\\color{@amber}{2x})$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"dydu"},{"color":"amber","style":"text-color","anchor":"dudx"}],"connections":[{"to":"formula-dydu","from":"dydu","color":"sky","label":"substitute","style":"solid"},{"to":"formula-dudx","from":"dudx","color":"amber","label":"substitute","style":"solid"}]},"order":2,"title":"Apply the chain rule","description":"We use the chain rule formula to combine our derivatives: $\\frac{dy}{dx} = \\frac{dy}{du} \\times \\frac{du}{dx}$."},{"type":"text","order":3,"title":"Substitute back and simplify","content":"To finish, we substitute $u = x^2 - 3$ back into our expression and simplify:\n\n$$\\begin{aligned} \\frac{dy}{dx} &= 5(x^2 - 3)^4 \\times 2x \\\\ &= 10x(x^2 - 3)^4 \\end{aligned}$$\n\nThis matches the expected derivative of $10x(x^2 - 3)^4$ in the markscheme, earning the method mark (M1) for applying the chain rule, and the accuracy mark (A1) for the correct final simplified expression."}]},{"id":"e6811698-1c26-47c5-82b0-bf8f153fe15d:binomial_expansion","label":"(b) Binomial Expansion and Term-by-Term Differentiation","steps":[{"type":"text","order":0,"title":"State the method","content":"We need to differentiate $(x^2 - 3)^5$ with respect to $x$.\n\nWhile this derivative is commonly found using the chain rule, we will solve it using **binomial expansion** to fully expand the expression first, followed by **term-by-term differentiation**."},{"type":"text","order":1,"title":"Apply Binomial Theorem","content":"Recall the binomial expansion formula:\n\n$$(a + b)^n = \\sum_{k=0}^{n} \\binom{n}{k} a^{n-k} b^k$$\n\nHere, we have $a = x^2$, $b = -3$, and $n = 5$. Let's write out the terms:\n\n$$\\begin{aligned} (x^2 - 3)^5 = \\;& \\binom{5}{0}(x^2)^5(-3)^0 + \\binom{5}{1}(x^2)^4(-3)^1 + \\binom{5}{2}(x^2)^3(-3)^2 \\\\ &+ \\binom{5}{3}(x^2)^2(-3)^3 + \\binom{5}{4}(x^2)^1(-3)^4 + \\binom{5}{5}(x^2)^0(-3)^5 \\end{aligned}$$"},{"type":"text","order":2,"title":"Simplify the expansion","content":"Now we calculate the coefficients and simplify each term using the binomial coefficients:\n* $\\binom{5}{0} = 1$, $\\binom{5}{1} = 5$, $\\binom{5}{2} = 10$, $\\binom{5}{3} = 10$, $\\binom{5}{4} = 5$, $\\binom{5}{5} = 1$\n\nSubstituting these values in, we get:\n\n$$\\begin{aligned} (x^2 - 3)^5 &= 1(x^{10})(1) + 5(x^8)(-3) + 10(x^6)(9) + 10(x^4)(-27) + 5(x^2)(81) + 1(1)(-243) \\\\ &= x^{10} - 15x^8 + 90x^6 - 270x^4 + 405x^2 - 243 \\end{aligned}$$"},{"type":"text","order":3,"title":"Differentiate term-by-term","content":"Now we apply the power rule for differentiation, $\\frac{d}{dx}(x^n) = nx^{n-1}$, to each term of the expanded polynomial:\n\n$$\\begin{aligned} \\frac{d}{dx}\\left[(x^2 - 3)^5\\right] &= \\frac{d}{dx}\\left(x^{10} - 15x^8 + 90x^6 - 270x^4 + 405x^2 - 243\\right) \\\\ &= 10x^9 - 15(8x^7) + 90(6x^5) - 270(4x^3) + 405(2x) - 0 \\\\ &= 10x^9 - 120x^7 + 540x^5 - 1080x^3 + 810x \\end{aligned}$$"},{"type":"text","order":4,"title":"Factorise the derivative","content":"To check our result against the standard chain rule form, let's factor out the common factor $10x$:\n\n$$\\frac{d}{dx}\\left[(x^2 - 3)^5\\right] = 10x\\left(x^8 - 12x^6 + 54x^4 - 108x^2 + 81\\right)$$\n\nNotice that the term in the brackets is the perfect fourth-power expansion of $(x^2 - 3)^4$:\n\n$$(x^2 - 3)^4 = x^8 - 12x^6 + 54x^4 - 108x^2 + 81$$\n\nSubstituting this back gives our final simplified answer:\n\n$$10x(x^2 - 3)^4$$"}]}],"generatedAt":"2026-06-09T02:01:26.090Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"1dac23ab-3f17-4825-b445-772fc700dfad","revisionId":"fea297ff-cf22-4871-8256-8ce7328d1520","questionNumber":"3","questionType":"LA","specification":"Differentiate with respect to $x$,","difficulty":1,"difficultyBand":"easy","tier":null,"totalMarks":4,"sourceTitle":"9.3 The Chain Rule","calculatorUse":null,"isCalculation":true,"parts":[{"id":"4d00d647-ced1-4326-8dce-5e3f14ef67fd","label":"a","content":"$$x + \\frac{5}{x^2 + 4}$","markscheme":"- Rewrites as $x + 5(x^2 + 4)^{-1}$ and differentiates to $1 - 5(x^2 + 4)^{-2} \\times (2x)$ M1\n- Final answer $1 - 10x(x^2 + 4)^{-2}$ A1","marks":2,"order":0,"markingRubric":{"totalMarks":2,"points":[{"id":"p1","marks":1,"description":"Differentiates $x$ to $1$ and applies chain rule to the fraction.","code":"M"},{"id":"p2","marks":1,"description":"Final answer $1 - 10x(x^2 + 4)^{-2}$.","code":"A"}],"finalAnswers":["1 - 10x(x^2 + 4)^{-2}"],"generatedAt":"2026-05-06T03:36:08.478Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"chainrule","label":"Chain Rule","steps":[{"type":"text","order":0,"title":"Rewrite the expression","content":"We rewrite the fraction using a negative exponent: $y = x + 5(x^2 + 4)^{-1}$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y = x + 5\\htmlData{anchor=outer}{\\color{@sky}{(x^2 + 4)^{-1}}}$"},{"latex":"$$\\frac{dy}{dx} = 1 + 5 \\cdot \\htmlData{anchor=diff-outer}{\\color{@sky}{-1(x^2 + 4)^{-2}}} \\cdot \\htmlData{anchor=diff-inner}{\\color{@amber}{\\frac{d}{dx}(x^2 + 4)}}$"}],"highlights":[{"color":"sky","style":"box","anchor":"outer"},{"color":"sky","style":"text-color","anchor":"diff-outer"},{"color":"amber","style":"underline","anchor":"diff-inner"}],"connections":[{"to":"diff-outer","from":"outer","color":"sky","label":"Power rule","style":"solid"}]},"order":1,"title":"Differentiate term-by-term","description":"We differentiate term-by-term. The derivative of $x$ is $1$. For $5(x^2 + 4)^{-1}$, we apply the chain rule."},{"type":"text","order":2,"title":"Apply Chain Rule","content":"Differentiating the inner function gives $\\frac{d}{dx}(x^2 + 4) = 2x$. Substituting this back: $\\frac{dy}{dx} = 1 - 5(x^2 + 4)^{-2} \\times (2x)$."},{"type":"text","order":3,"title":"Simplify final answer","content":"Simplify by multiplying the coefficients: $-5 \\times 2x = -10x$. The final derivative is: $\\frac{dy}{dx} = 1 - 10x(x^2 + 4)^{-2}$."}]},{"id":"quotientrule","label":"Quotient Rule","steps":[{"type":"text","order":0,"title":"Break down the terms","content":"We differentiate term-by-term: $\\frac{dy}{dx} = \\frac{d}{dx}(x) + \\frac{d}{dx}( \\frac{5}{x^2+4} )$."},{"type":"text","order":1,"title":"Differentiate the first term","content":"The derivative of the first term $x$ is $1$."},{"type":"hyper_latex","block":{"nodes":[{"text":"For the second term, we apply the quotient rule:"},{"latex":"$$\\frac{d}{dx}( \\frac{u}{v} ) = \\frac{u'v - uv'}{v^2}$"},{"text":"Let:"},{"latex":"$$\\htmlData{anchor=u-term}{\\color{@sky}{u = 5}} \\implies \\htmlData{anchor=du-term}{\\color{@sky}{u' = 0}}$"},{"latex":"$$\\htmlData{anchor=v-term}{\\color{@amber}{v = x^2 + 4}} \\implies \\htmlData{anchor=dv-term}{\\color{@amber}{v' = 2x}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-term"},{"color":"amber","style":"text-color","anchor":"v-term"}],"connections":[{"to":"du-term","from":"u-term","color":"sky","label":"differentiate","style":"solid"},{"to":"dv-term","from":"v-term","color":"amber","label":"differentiate","style":"solid"}]},"order":2,"title":"Apply the Quotient Rule","description":"Identify the numerator as $u$ and the denominator as $v$, then find their respective derivatives."},{"type":"text","order":3,"title":"Substitute into the formula","content":"Now we substitute these derivatives into the quotient rule formula: $\\frac{d}{dx}( \\frac{5}{x^2+4} ) = \\frac{(0)(x^2+4) - (5)(2x)}{(x^2+4)^2}$. Simplifying the numerator: $\\frac{d}{dx}( \\frac{5}{x^2+4} ) = \\frac{-10x}{(x^2+4)^2} = -10x(x^2+4)^{-2}$."},{"type":"text","order":4,"title":"State the final answer","content":"Combine the derivatives of both terms to get the complete derivative: $\\frac{dy}{dx} = 1 - 10x(x^2 + 4)^{-2}$."}]}],"generatedAt":"2026-06-09T03:57:04.679Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"98a15c11-9a23-4933-bd98-01ee0080e3e8","label":"b","content":"$$8 + e^{x^2}$","markscheme":"- Differentiates $e^{x^2}$ to $2x e^{x^2}$ M1\n- Final answer $2x e^{x^2}$ A1","marks":2,"order":1,"markingRubric":{"totalMarks":2,"points":[{"id":"p1","marks":1,"description":"Applies chain rule to $e^{x^2}$ to get $e^{x^2} \\times 2x$.","code":"M"},{"id":"p2","marks":1,"description":"Final answer $2x e^{x^2}$.","code":"A"}],"finalAnswers":["2x e^{x^2}"],"generatedAt":"2026-05-06T03:36:08.478Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"direct_chain_rule","label":"Direct Chain Rule","steps":[{"type":"text","order":0,"title":"Identify the terms","content":"We want to differentiate the expression $8 + e^{x^2}$ with respect to $x$. Since differentiation is a linear operation, we can differentiate each part of the expression individually:\n\n$$\\frac{d}{dx}(8 + e^{x^2}) = \\frac{d}{dx}(8) + \\frac{d}{dx}(e^{x^2})$$"},{"type":"text","order":1,"title":"Differentiate the constant","content":"The first term is the constant $8$. Since constants do not change with respect to $x$, their derivative is always $0$:\n\n$$\\frac{d}{dx}(8) = 0$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Exponent: } \\htmlData{anchor=u}{\\color{@sky}{x^2}} \\quad \\xrightarrow{\\text{Derivative}} \\quad \\htmlData{anchor=du}{\\color{@amber}{2x}}$"},{"latex":"$$\\frac{d}{dx}\\left( e^{\\htmlData{anchor=u-sub}{\\color{@sky}{x^2}}} \\right) = \\htmlData{anchor=du-sub}{\\color{@amber}{2x}} e^{\\color{@sky}{x^2}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u"},{"color":"amber","style":"text-color","anchor":"du"}],"connections":[{"to":"u-sub","from":"u","color":"sky","label":"exponent","style":"solid"},{"to":"du-sub","from":"du","color":"amber","label":"f'(x)","style":"solid"}]},"order":2,"title":"Apply the chain rule","description":"For the exponential term, we apply the chain rule directly. The derivative of $e^{f(x)}$ is simply the derivative of the exponent, $f'(x)$, multiplied by the original exponential term:"},{"type":"text","order":3,"title":"Combine the results","content":"Now we simply add the two derivatives together to get our final result:\n\n$$\\frac{d}{dx}(8 + e^{x^2}) = 0 + 2x e^{x^2} = 2x e^{x^2}$$\n\nThis aligns perfectly with the exam marking scheme:\n- You earn **M1** for correctly differentiating the exponential term to $2x e^{x^2}$.\n- You earn **A1** for presenting the final simplified derivative."}]},{"id":"substitution_chain_rule","label":"Chain Rule with Substitution","steps":[{"type":"text","order":0,"title":"Define the substitution","content":"To differentiate the function $y = 8 + e^{x^2}$ using the chain rule with substitution, we first identify the inner function and replace it with a new variable $u$.\n\nLet the inner function be:\n$$u = x^2$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y = 8 + e^{\\htmlData{anchor=u-var}{\\color{@sky}{x^2}}}$"},{"latex":"$$y = 8 + e^{\\htmlData{anchor=u-sub}{\\color{@sky}{u}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-var"},{"color":"sky","style":"text-color","anchor":"u-sub"}],"connections":[{"to":"u-sub","from":"u-var","color":"sky","label":null,"style":"solid"}]},"order":1,"title":"Express y in terms of u","description":"Substitute $u$ into our original function to express $y$ in terms of $u$."},{"type":"text","order":2,"title":"Find the derivatives","content":"Now, we differentiate both equations:\n\n1. Differentiate $u = x^2$ with respect to $x$:\n$$\\frac{du}{dx} = 2x$$\n\n2. Differentiate $y = 8 + e^u$ with respect to $u$. Note that the constant $8$ differentiates to $0$, and the derivative of $e^u$ is $e^u$:\n$$\\frac{dy}{du} = e^u$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\frac{dy}{dx} = \\htmlData{anchor=dydu-form}{\\color{@pink}{\\frac{dy}{du}}} \\cdot \\htmlData{anchor=dudx-form}{\\color{@purple}{\\frac{du}{dx}}}$"},{"latex":"$$\\frac{dy}{dx} = \\htmlData{anchor=dydu-sub}{\\color{@pink}{e^u}} \\cdot \\htmlData{anchor=dudx-sub}{\\color{@purple}{2x}}$"}],"highlights":[{"color":"pink","style":"text-color","anchor":"dydu-form"},{"color":"purple","style":"text-color","anchor":"dudx-form"},{"color":"pink","style":"text-color","anchor":"dydu-sub"},{"color":"purple","style":"text-color","anchor":"dudx-sub"}],"connections":[{"to":"dydu-sub","from":"dydu-form","color":"pink","label":"substitute","style":"solid"},{"to":"dudx-sub","from":"dudx-form","color":"purple","label":"substitute","style":"solid"}]},"order":3,"title":"Apply the Chain Rule","description":"Now, apply the chain rule formula to relate our derivatives."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\frac{dy}{dx} = 2x e^{\\htmlData{anchor=u-back-sub}{\\color{@sky}{u}}}$"},{"latex":"$$\\frac{dy}{dx} = 2x e^{\\htmlData{anchor=u-back-val}{\\color{@sky}{x^2}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-back-sub"},{"color":"sky","style":"text-color","anchor":"u-back-val"}],"connections":[{"to":"u-back-val","from":"u-back-sub","color":"sky","label":"u = x^2","style":"solid"}]},"order":4,"title":"Substitute back to x","description":"Substitute $u = x^2$ back into the expression to obtain the final derivative in terms of $x$."},{"type":"text","order":5,"title":"State final answer","content":"The final derivative of the function is:\n\n$$2x e^{x^2}$$\n\nThis earns **1 method mark** (M1) for correctly differentiating $e^{x^2}$ to $2x e^{x^2}$, and **1 accuracy mark** (A1) for the final correct expression."}]}],"generatedAt":"2026-06-09T03:58:50.026Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"4d00d647-ced1-4326-8dce-5e3f14ef67fd:chainrule","label":"(a) Chain Rule","steps":[{"type":"text","order":0,"title":"Rewrite the expression","content":"We rewrite the fraction using a negative exponent: $y = x + 5(x^2 + 4)^{-1}$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y = x + 5\\htmlData{anchor=outer}{\\color{@sky}{(x^2 + 4)^{-1}}}$"},{"latex":"$$\\frac{dy}{dx} = 1 + 5 \\cdot \\htmlData{anchor=diff-outer}{\\color{@sky}{-1(x^2 + 4)^{-2}}} \\cdot \\htmlData{anchor=diff-inner}{\\color{@amber}{\\frac{d}{dx}(x^2 + 4)}}$"}],"highlights":[{"color":"sky","style":"box","anchor":"outer"},{"color":"sky","style":"text-color","anchor":"diff-outer"},{"color":"amber","style":"underline","anchor":"diff-inner"}],"connections":[{"to":"diff-outer","from":"outer","color":"sky","label":"Power rule","style":"solid"}]},"order":1,"title":"Differentiate term-by-term","description":"We differentiate term-by-term. The derivative of $x$ is $1$. For $5(x^2 + 4)^{-1}$, we apply the chain rule."},{"type":"text","order":2,"title":"Apply Chain Rule","content":"Differentiating the inner function gives $\\frac{d}{dx}(x^2 + 4) = 2x$. Substituting this back: $\\frac{dy}{dx} = 1 - 5(x^2 + 4)^{-2} \\times (2x)$."},{"type":"text","order":3,"title":"Simplify final answer","content":"Simplify by multiplying the coefficients: $-5 \\times 2x = -10x$. The final derivative is: $\\frac{dy}{dx} = 1 - 10x(x^2 + 4)^{-2}$."}]},{"id":"4d00d647-ced1-4326-8dce-5e3f14ef67fd:quotientrule","label":"(a) Quotient Rule","steps":[{"type":"text","order":0,"title":"Break down the terms","content":"We differentiate term-by-term: $\\frac{dy}{dx} = \\frac{d}{dx}(x) + \\frac{d}{dx}( \\frac{5}{x^2+4} )$."},{"type":"text","order":1,"title":"Differentiate the first term","content":"The derivative of the first term $x$ is $1$."},{"type":"hyper_latex","block":{"nodes":[{"text":"For the second term, we apply the quotient rule:"},{"latex":"$$\\frac{d}{dx}( \\frac{u}{v} ) = \\frac{u'v - uv'}{v^2}$"},{"text":"Let:"},{"latex":"$$\\htmlData{anchor=u-term}{\\color{@sky}{u = 5}} \\implies \\htmlData{anchor=du-term}{\\color{@sky}{u' = 0}}$"},{"latex":"$$\\htmlData{anchor=v-term}{\\color{@amber}{v = x^2 + 4}} \\implies \\htmlData{anchor=dv-term}{\\color{@amber}{v' = 2x}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-term"},{"color":"amber","style":"text-color","anchor":"v-term"}],"connections":[{"to":"du-term","from":"u-term","color":"sky","label":"differentiate","style":"solid"},{"to":"dv-term","from":"v-term","color":"amber","label":"differentiate","style":"solid"}]},"order":2,"title":"Apply the Quotient Rule","description":"Identify the numerator as $u$ and the denominator as $v$, then find their respective derivatives."},{"type":"text","order":3,"title":"Substitute into the formula","content":"Now we substitute these derivatives into the quotient rule formula: $\\frac{d}{dx}( \\frac{5}{x^2+4} ) = \\frac{(0)(x^2+4) - (5)(2x)}{(x^2+4)^2}$. Simplifying the numerator: $\\frac{d}{dx}( \\frac{5}{x^2+4} ) = \\frac{-10x}{(x^2+4)^2} = -10x(x^2+4)^{-2}$."},{"type":"text","order":4,"title":"State the final answer","content":"Combine the derivatives of both terms to get the complete derivative: $\\frac{dy}{dx} = 1 - 10x(x^2 + 4)^{-2}$."}]},{"id":"98a15c11-9a23-4933-bd98-01ee0080e3e8:direct_chain_rule","label":"(b) Direct Chain Rule","steps":[{"type":"text","order":0,"title":"Identify the terms","content":"We want to differentiate the expression $8 + e^{x^2}$ with respect to $x$. Since differentiation is a linear operation, we can differentiate each part of the expression individually:\n\n$$\\frac{d}{dx}(8 + e^{x^2}) = \\frac{d}{dx}(8) + \\frac{d}{dx}(e^{x^2})$$"},{"type":"text","order":1,"title":"Differentiate the constant","content":"The first term is the constant $8$. Since constants do not change with respect to $x$, their derivative is always $0$:\n\n$$\\frac{d}{dx}(8) = 0$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Exponent: } \\htmlData{anchor=u}{\\color{@sky}{x^2}} \\quad \\xrightarrow{\\text{Derivative}} \\quad \\htmlData{anchor=du}{\\color{@amber}{2x}}$"},{"latex":"$$\\frac{d}{dx}\\left( e^{\\htmlData{anchor=u-sub}{\\color{@sky}{x^2}}} \\right) = \\htmlData{anchor=du-sub}{\\color{@amber}{2x}} e^{\\color{@sky}{x^2}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u"},{"color":"amber","style":"text-color","anchor":"du"}],"connections":[{"to":"u-sub","from":"u","color":"sky","label":"exponent","style":"solid"},{"to":"du-sub","from":"du","color":"amber","label":"f'(x)","style":"solid"}]},"order":2,"title":"Apply the chain rule","description":"For the exponential term, we apply the chain rule directly. The derivative of $e^{f(x)}$ is simply the derivative of the exponent, $f'(x)$, multiplied by the original exponential term:"},{"type":"text","order":3,"title":"Combine the results","content":"Now we simply add the two derivatives together to get our final result:\n\n$$\\frac{d}{dx}(8 + e^{x^2}) = 0 + 2x e^{x^2} = 2x e^{x^2}$$\n\nThis aligns perfectly with the exam marking scheme:\n- You earn **M1** for correctly differentiating the exponential term to $2x e^{x^2}$.\n- You earn **A1** for presenting the final simplified derivative."}]},{"id":"98a15c11-9a23-4933-bd98-01ee0080e3e8:substitution_chain_rule","label":"(b) Chain Rule with Substitution","steps":[{"type":"text","order":0,"title":"Define the substitution","content":"To differentiate the function $y = 8 + e^{x^2}$ using the chain rule with substitution, we first identify the inner function and replace it with a new variable $u$.\n\nLet the inner function be:\n$$u = x^2$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y = 8 + e^{\\htmlData{anchor=u-var}{\\color{@sky}{x^2}}}$"},{"latex":"$$y = 8 + e^{\\htmlData{anchor=u-sub}{\\color{@sky}{u}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-var"},{"color":"sky","style":"text-color","anchor":"u-sub"}],"connections":[{"to":"u-sub","from":"u-var","color":"sky","label":null,"style":"solid"}]},"order":1,"title":"Express y in terms of u","description":"Substitute $u$ into our original function to express $y$ in terms of $u$."},{"type":"text","order":2,"title":"Find the derivatives","content":"Now, we differentiate both equations:\n\n1. Differentiate $u = x^2$ with respect to $x$:\n$$\\frac{du}{dx} = 2x$$\n\n2. Differentiate $y = 8 + e^u$ with respect to $u$. Note that the constant $8$ differentiates to $0$, and the derivative of $e^u$ is $e^u$:\n$$\\frac{dy}{du} = e^u$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\frac{dy}{dx} = \\htmlData{anchor=dydu-form}{\\color{@pink}{\\frac{dy}{du}}} \\cdot \\htmlData{anchor=dudx-form}{\\color{@purple}{\\frac{du}{dx}}}$"},{"latex":"$$\\frac{dy}{dx} = \\htmlData{anchor=dydu-sub}{\\color{@pink}{e^u}} \\cdot \\htmlData{anchor=dudx-sub}{\\color{@purple}{2x}}$"}],"highlights":[{"color":"pink","style":"text-color","anchor":"dydu-form"},{"color":"purple","style":"text-color","anchor":"dudx-form"},{"color":"pink","style":"text-color","anchor":"dydu-sub"},{"color":"purple","style":"text-color","anchor":"dudx-sub"}],"connections":[{"to":"dydu-sub","from":"dydu-form","color":"pink","label":"substitute","style":"solid"},{"to":"dudx-sub","from":"dudx-form","color":"purple","label":"substitute","style":"solid"}]},"order":3,"title":"Apply the Chain Rule","description":"Now, apply the chain rule formula to relate our derivatives."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\frac{dy}{dx} = 2x e^{\\htmlData{anchor=u-back-sub}{\\color{@sky}{u}}}$"},{"latex":"$$\\frac{dy}{dx} = 2x e^{\\htmlData{anchor=u-back-val}{\\color{@sky}{x^2}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-back-sub"},{"color":"sky","style":"text-color","anchor":"u-back-val"}],"connections":[{"to":"u-back-val","from":"u-back-sub","color":"sky","label":"u = x^2","style":"solid"}]},"order":4,"title":"Substitute back to x","description":"Substitute $u = x^2$ back into the expression to obtain the final derivative in terms of $x$."},{"type":"text","order":5,"title":"State final answer","content":"The final derivative of the function is:\n\n$$2x e^{x^2}$$\n\nThis earns **1 method mark** (M1) for correctly differentiating $e^{x^2}$ to $2x e^{x^2}$, and **1 accuracy mark** (A1) for the final correct expression."}]}],"generatedAt":"2026-06-09T03:59:03.019Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"002faeea-0afa-4ba5-a773-8c786542dd31","revisionId":"837e9dd4-1c55-486e-b527-8c96e565669e","questionNumber":null,"questionType":"LA","specification":"Differentiate with respect to $x$,","difficulty":1,"difficultyBand":"easy","tier":null,"totalMarks":4,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"dba1b6b8-662c-47f5-a158-7dbb59981ef2","label":"a","content":"$$4x^3 + e^{6x}$","markscheme":"- Differentiates $4x^3$ to $12x^2$ B1\n- Differentiates $e^{6x}$ to $6e^{6x}$ B1","marks":2,"order":0,"markingRubric":{"totalMarks":2,"points":[{"id":"p1","marks":1,"description":"Differentiates $4x^3$ to obtain $12x^2$","code":"B"},{"id":"p2","marks":1,"description":"Differentiates $e^{6x}$ to obtain $6e^{6x}$","code":"B"}],"finalAnswers":["12x^2 + 6e^{6x}"],"generatedAt":"2026-05-06T14:03:52.912Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"term_by_term_differentiation","label":"Term-by-Term Differentiation","steps":[{"type":"text","order":0,"title":"Identify the approach","content":"To differentiate the expression $4x^3 + e^{6x}$ with respect to $x$, we can use the sum rule of differentiation. This rule states that we can differentiate each term individually:\n\n$$\\frac{d}{dx} [f(x) + g(x)] = \\frac{d}{dx}[f(x)] + \\frac{d}{dx}[g(x)]$$\n\nLet's break this down into two separate parts:\n1. Differentiate the algebraic term $4x^3$.\n2. Differentiate the exponential term $e^{6x}$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\frac{d}{dx}(\\htmlData{anchor=first-term}{\\color{@sky}{4x^3}}) = 4 \\cdot \\htmlData{anchor=power-rule}{\\color{@amber}{3x^{3-1}}}$"},{"latex":"$$\\phantom{\\frac{d}{dx}(4x^3)} = \\htmlData{anchor=first-deriv}{\\color{@purple}{12x^2}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"first-term"},{"color":"amber","style":"text-color","anchor":"power-rule"},{"color":"purple","style":"text-color","anchor":"first-deriv"}],"connections":[{"to":"power-rule","from":"first-term","color":"amber","label":"power rule","style":"solid"},{"to":"first-deriv","from":"power-rule","color":"purple","label":"simplify","style":"solid"}]},"order":1,"title":"Differentiate the first term","description":"Apply the power rule, $\\frac{d}{dx}(x^n) = n x^{n-1}$, to the first term $4x^3$. Multiplying the exponent by the coefficient gives $4 \\times 3 = 12$, and we reduce the exponent by $1$ to get $x^2$. This earns the first mark."},{"type":"hyper_latex","block":{"nodes":[{"text":"Using the chain rule, $\\frac{d}{dx}(e^{kx}) = k e^{kx}$:"},{"latex":"$$\\frac{d}{dx}(\\htmlData{anchor=second-term}{\\color{@sky}{e^{6x}}}) = \\htmlData{anchor=chain-rule}{\\color{@amber}{6}} \\cdot e^{\\htmlData{anchor=exponent}{\\color{@purple}{6x}}}$"},{"latex":"$$\\phantom{\\frac{d}{dx}(e^{6x})} = \\htmlData{anchor=second-deriv}{\\color{@teal}{6e^{6x}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"second-term"},{"color":"amber","style":"text-color","anchor":"chain-rule"},{"color":"teal","style":"text-color","anchor":"second-deriv"}],"connections":[{"to":"chain-rule","from":"second-term","color":"amber","label":"multiply by derivative of 6x","style":"solid"},{"to":"second-deriv","from":"exponent","color":"teal","label":"exponent remains unchanged","style":"solid"}]},"order":2,"title":"Differentiate the second term","description":"Apply the chain rule to differentiate the exponential term $e^{6x}$. The derivative of $e^{u}$ with respect to $u$ is $e^{u}$, and the derivative of $6x$ with respect to $x$ is $6$. This earns the second mark."},{"type":"text","order":3,"title":"Combine the results","content":"Now, we sum our two individual derivatives together to get the final result for Part 1:\n\n$$\\frac{d}{dx}(4x^3 + e^{6x}) = 12x^2 + 6e^{6x}$$"},{"type":"text","order":4,"title":"Differentiate Part 2","content":"For Part 2, we need to differentiate $(x^2 + 10)^3$ with respect to $x$. Since this is a composite function of the form $[f(x)]^n$, we apply the chain rule:\n\n$$\\frac{d}{dx}([f(x)]^n) = n[f(x)]^{n-1} \\cdot f'(x)$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\frac{d}{dx}(\\htmlData{anchor=p2-orig}{\\color{@sky}{\\left(x^2 + 10\\right)^3}}) = \\htmlData{anchor=p2-outer}{\\color{@amber}{3\\left(x^2 + 10\\right)^2}} \\cdot \\frac{d}{dx}(\\htmlData{anchor=p2-inner}{\\color{@purple}{x^2 + 10}})$"},{"latex":"$$\\phantom{\\frac{d}{dx}((x^2 + 10)^3)} = 3\\left(x^2 + 10\\right)^2 \\cdot \\htmlData{anchor=p2-inner-deriv}{\\color{@purple}{2x}}$"},{"latex":"$$\\phantom{\\frac{d}{dx}((x^2 + 10)^3)} = \\htmlData{anchor=p2-final}{\\color{@teal}{6x\\left(x^2 + 10\\right)^2}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"p2-orig"},{"color":"amber","style":"text-color","anchor":"p2-outer"},{"color":"purple","style":"text-color","anchor":"p2-inner-deriv"},{"color":"teal","style":"text-color","anchor":"p2-final"}],"connections":[{"to":"p2-inner-deriv","from":"p2-inner","color":"purple","label":"differentiate inner function","style":"solid"},{"to":"p2-final","from":"p2-inner-deriv","color":"teal","label":"multiply and simplify","style":"solid"}]},"order":5,"title":"Apply the chain rule","description":"Differentiate the outer power function first, keeping the inner expression the same, then multiply by the derivative of the inner function $x^2 + 10$."}]}],"generatedAt":"2026-06-09T02:57:49.346Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"a0081eec-278f-408c-bd4a-0ea5652e7338","label":"b","content":"$$(x^2 + 10)^3$","markscheme":"- Applies chain rule: $3(x^2 + 10)^2 \\times \\dots$ M1\n- Correct derivative $6x(x^2 + 10)^2$ A1","marks":2,"order":1,"markingRubric":{"totalMarks":2,"points":[{"id":"p1","marks":1,"description":"Evidence of chain rule application, e.g., $3(x^2 + 10)^2$","code":"M"},{"id":"p2","marks":1,"description":"Final derivative $6x(x^2 + 10)^2$ or equivalent simplified form","code":"A"}],"finalAnswers":["6x(x^2 + 10)^2"],"generatedAt":"2026-05-06T14:03:52.912Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"chain_rule","label":"Chain Rule","steps":[{"type":"text","order":0,"title":"Identify the chain rule","content":"To differentiate the function $(x^2 + 10)^3$ with respect to $x$, we apply the **Chain Rule**.\n\nThe Chain Rule states that for a composite function $y = f(g(x))$:\n\n$$\\frac{dy}{dx} = f'(g(x)) \\times g'(x)$$\n\nIn plain terms, we differentiate the \"outer\" function with respect to the \"inner\" function, and then multiply by the derivative of the \"inner\" function."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y = (\\htmlData{anchor=inner-fx}{\\color{@sky}{x^2 + 10}})^{\\htmlData{anchor=outer-power}{\\color{@pink}{3}}}$"},{"latex":"$$3(\\htmlData{anchor=inner-copy}{\\color{@sky}{x^2 + 10}})^{\\htmlData{anchor=new-power}{\\color{@pink}{2}}} \\times \\dots$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"inner-fx"},{"color":"sky","style":"text-color","anchor":"inner-copy"},{"color":"pink","style":"text-color","anchor":"outer-power"},{"color":"pink","style":"text-color","anchor":"new-power"}],"connections":[{"to":"new-power","from":"outer-power","color":"pink","label":"power rule","style":"solid"}]},"order":1,"title":"Differentiate the outer function","description":"Treating the inner part $x^2 + 10$ as a single variable, we use the power rule on the outer power $3$ to get $3(x^2 + 10)^2$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Inner function: } \\htmlData{anchor=inner-only}{\\color{@sky}{x^2 + 10}}$"},{"latex":"$$\\text{Derivative: } \\frac{d}{dx}(\\color{@sky}{x^2 + 10}) = \\htmlData{anchor=inner-deriv}{\\color{@amber}{2x}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"inner-only"},{"color":"amber","style":"text-color","anchor":"inner-deriv"}],"connections":[{"to":"inner-deriv","from":"inner-only","color":"amber","label":"differentiate","style":"solid"}]},"order":2,"title":"Differentiate the inner function","description":"Next, we find the derivative of the inner function with respect to $x$."},{"type":"text","order":3,"title":"Multiply the derivatives","content":"Now, we multiply the derivative of the outer function by the derivative of the inner function:\n\n$$\\frac{d}{dx}(x^2 + 10)^3 = 3(x^2 + 10)^2 \\times (2x)$$\n\nApplying this step correctly earns you the first method mark **(M1)**."},{"type":"text","order":4,"title":"Simplify the final answer","content":"Simplify the expression by multiplying the coefficients:\n\n$$3 \\times 2x = 6x$$\n\nThus, the final derivative is:\n\n$$6x(x^2 + 10)^2$$\n\nThis completes the calculation and earns you the final accuracy mark **(A1)**."}]},{"id":"algebraic_expansion","label":"Expansion and Term-by-Term Differentiation","steps":[{"type":"text","order":0,"title":"Understand the method","content":"To differentiate the function $y = (x^2 + 10)^3$ using the **Expansion and Term-by-Term Differentiation** method, we will first fully expand the cubic expression algebraically. Once we have a polynomial in terms of powers of $x$, we can differentiate each term individually using the power rule:\n\n$$\\frac{d}{dx}(x^n) = nx^{n-1}$$"},{"type":"text","order":1,"title":"Expand the expression","content":"We can expand the expression $(x^2 + 10)^3$ using the binomial expansion formula for a cubic expression:\n\n$$(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$$\n\nSubstituting $a = x^2$ and $b = 10$, we get:\n\n$$\\begin{aligned} (x^2 + 10)^3 &= (x^2)^3 + 3(x^2)^2(10) + 3(x^2)(10)^2 + 10^3 \\\\ &= x^6 + 3(x^4)(10) + 3(x^2)(100) + 1000 \\\\ &= x^6 + 30x^4 + 300x^2 + 1000 \\end{aligned}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y = \\htmlData{anchor=term1}{\\color{@sky}{x^6}} + \\htmlData{anchor=term2}{\\color{@amber}{30x^4}} + \\htmlData{anchor=term3}{\\color{@purple}{300x^2}} + \\htmlData{anchor=term4}{\\color{@teal}{1000}}$"},{"latex":"$$\\frac{dy}{dx} = \\htmlData{anchor=diff1}{\\color{@sky}{6x^5}} + \\htmlData{anchor=diff2}{\\color{@amber}{120x^3}} + \\htmlData{anchor=diff3}{\\color{@purple}{600x}} + \\htmlData{anchor=diff4}{\\color{@teal}{0}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"term1"},{"color":"amber","style":"text-color","anchor":"term2"},{"color":"purple","style":"text-color","anchor":"term3"},{"color":"teal","style":"text-color","anchor":"term4"}],"connections":[{"to":"diff1","from":"term1","color":"sky","label":"Power Rule","style":"solid"},{"to":"diff2","from":"term2","color":"amber","label":"Power Rule","style":"solid"},{"to":"diff3","from":"term3","color":"purple","label":"Power Rule","style":"solid"},{"to":"diff4","from":"term4","color":"teal","label":"Constant Rule","style":"solid"}]},"order":2,"title":"Differentiate term-by-term","description":"Now, we apply the power rule to differentiate each term of the expanded polynomial with respect to $x$ individually."},{"type":"text","order":3,"title":"Factor the final answer","content":"To simplify our derivative and match it with the standard factored form, we can factor out the greatest common factor, which is $6x$:\n\n$$\\begin{aligned} \\frac{dy}{dx} &= 6x^5 + 120x^3 + 600x \\\\ &= 6x(x^4 + 20x^2 + 100) \\end{aligned}$$\n\nNotice that the quadratic expression inside the brackets is a perfect square trinomial:\n\n$$x^4 + 20x^2 + 100 = (x^2 + 10)^2$$\n\nSo, the fully factored derivative is:\n\n$$\\frac{dy}{dx} = 6x(x^2 + 10)^2$$"}]}],"generatedAt":"2026-06-09T02:59:07.093Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"dba1b6b8-662c-47f5-a158-7dbb59981ef2:term_by_term_differentiation","label":"(a) Term-by-Term Differentiation","steps":[{"type":"text","order":0,"title":"Identify the approach","content":"To differentiate the expression $4x^3 + e^{6x}$ with respect to $x$, we can use the sum rule of differentiation. This rule states that we can differentiate each term individually:\n\n$$\\frac{d}{dx} [f(x) + g(x)] = \\frac{d}{dx}[f(x)] + \\frac{d}{dx}[g(x)]$$\n\nLet's break this down into two separate parts:\n1. Differentiate the algebraic term $4x^3$.\n2. Differentiate the exponential term $e^{6x}$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\frac{d}{dx}(\\htmlData{anchor=first-term}{\\color{@sky}{4x^3}}) = 4 \\cdot \\htmlData{anchor=power-rule}{\\color{@amber}{3x^{3-1}}}$"},{"latex":"$$\\phantom{\\frac{d}{dx}(4x^3)} = \\htmlData{anchor=first-deriv}{\\color{@purple}{12x^2}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"first-term"},{"color":"amber","style":"text-color","anchor":"power-rule"},{"color":"purple","style":"text-color","anchor":"first-deriv"}],"connections":[{"to":"power-rule","from":"first-term","color":"amber","label":"power rule","style":"solid"},{"to":"first-deriv","from":"power-rule","color":"purple","label":"simplify","style":"solid"}]},"order":1,"title":"Differentiate the first term","description":"Apply the power rule, $\\frac{d}{dx}(x^n) = n x^{n-1}$, to the first term $4x^3$. Multiplying the exponent by the coefficient gives $4 \\times 3 = 12$, and we reduce the exponent by $1$ to get $x^2$. This earns the first mark."},{"type":"hyper_latex","block":{"nodes":[{"text":"Using the chain rule, $\\frac{d}{dx}(e^{kx}) = k e^{kx}$:"},{"latex":"$$\\frac{d}{dx}(\\htmlData{anchor=second-term}{\\color{@sky}{e^{6x}}}) = \\htmlData{anchor=chain-rule}{\\color{@amber}{6}} \\cdot e^{\\htmlData{anchor=exponent}{\\color{@purple}{6x}}}$"},{"latex":"$$\\phantom{\\frac{d}{dx}(e^{6x})} = \\htmlData{anchor=second-deriv}{\\color{@teal}{6e^{6x}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"second-term"},{"color":"amber","style":"text-color","anchor":"chain-rule"},{"color":"teal","style":"text-color","anchor":"second-deriv"}],"connections":[{"to":"chain-rule","from":"second-term","color":"amber","label":"multiply by derivative of 6x","style":"solid"},{"to":"second-deriv","from":"exponent","color":"teal","label":"exponent remains unchanged","style":"solid"}]},"order":2,"title":"Differentiate the second term","description":"Apply the chain rule to differentiate the exponential term $e^{6x}$. The derivative of $e^{u}$ with respect to $u$ is $e^{u}$, and the derivative of $6x$ with respect to $x$ is $6$. This earns the second mark."},{"type":"text","order":3,"title":"Combine the results","content":"Now, we sum our two individual derivatives together to get the final result for Part 1:\n\n$$\\frac{d}{dx}(4x^3 + e^{6x}) = 12x^2 + 6e^{6x}$$"},{"type":"text","order":4,"title":"Differentiate Part 2","content":"For Part 2, we need to differentiate $(x^2 + 10)^3$ with respect to $x$. Since this is a composite function of the form $[f(x)]^n$, we apply the chain rule:\n\n$$\\frac{d}{dx}([f(x)]^n) = n[f(x)]^{n-1} \\cdot f'(x)$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\frac{d}{dx}(\\htmlData{anchor=p2-orig}{\\color{@sky}{\\left(x^2 + 10\\right)^3}}) = \\htmlData{anchor=p2-outer}{\\color{@amber}{3\\left(x^2 + 10\\right)^2}} \\cdot \\frac{d}{dx}(\\htmlData{anchor=p2-inner}{\\color{@purple}{x^2 + 10}})$"},{"latex":"$$\\phantom{\\frac{d}{dx}((x^2 + 10)^3)} = 3\\left(x^2 + 10\\right)^2 \\cdot \\htmlData{anchor=p2-inner-deriv}{\\color{@purple}{2x}}$"},{"latex":"$$\\phantom{\\frac{d}{dx}((x^2 + 10)^3)} = \\htmlData{anchor=p2-final}{\\color{@teal}{6x\\left(x^2 + 10\\right)^2}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"p2-orig"},{"color":"amber","style":"text-color","anchor":"p2-outer"},{"color":"purple","style":"text-color","anchor":"p2-inner-deriv"},{"color":"teal","style":"text-color","anchor":"p2-final"}],"connections":[{"to":"p2-inner-deriv","from":"p2-inner","color":"purple","label":"differentiate inner function","style":"solid"},{"to":"p2-final","from":"p2-inner-deriv","color":"teal","label":"multiply and simplify","style":"solid"}]},"order":5,"title":"Apply the chain rule","description":"Differentiate the outer power function first, keeping the inner expression the same, then multiply by the derivative of the inner function $x^2 + 10$."}]},{"id":"a0081eec-278f-408c-bd4a-0ea5652e7338:chain_rule","label":"(b) Chain Rule","steps":[{"type":"text","order":0,"title":"Identify the chain rule","content":"To differentiate the function $(x^2 + 10)^3$ with respect to $x$, we apply the **Chain Rule**.\n\nThe Chain Rule states that for a composite function $y = f(g(x))$:\n\n$$\\frac{dy}{dx} = f'(g(x)) \\times g'(x)$$\n\nIn plain terms, we differentiate the \"outer\" function with respect to the \"inner\" function, and then multiply by the derivative of the \"inner\" function."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y = (\\htmlData{anchor=inner-fx}{\\color{@sky}{x^2 + 10}})^{\\htmlData{anchor=outer-power}{\\color{@pink}{3}}}$"},{"latex":"$$3(\\htmlData{anchor=inner-copy}{\\color{@sky}{x^2 + 10}})^{\\htmlData{anchor=new-power}{\\color{@pink}{2}}} \\times \\dots$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"inner-fx"},{"color":"sky","style":"text-color","anchor":"inner-copy"},{"color":"pink","style":"text-color","anchor":"outer-power"},{"color":"pink","style":"text-color","anchor":"new-power"}],"connections":[{"to":"new-power","from":"outer-power","color":"pink","label":"power rule","style":"solid"}]},"order":1,"title":"Differentiate the outer function","description":"Treating the inner part $x^2 + 10$ as a single variable, we use the power rule on the outer power $3$ to get $3(x^2 + 10)^2$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Inner function: } \\htmlData{anchor=inner-only}{\\color{@sky}{x^2 + 10}}$"},{"latex":"$$\\text{Derivative: } \\frac{d}{dx}(\\color{@sky}{x^2 + 10}) = \\htmlData{anchor=inner-deriv}{\\color{@amber}{2x}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"inner-only"},{"color":"amber","style":"text-color","anchor":"inner-deriv"}],"connections":[{"to":"inner-deriv","from":"inner-only","color":"amber","label":"differentiate","style":"solid"}]},"order":2,"title":"Differentiate the inner function","description":"Next, we find the derivative of the inner function with respect to $x$."},{"type":"text","order":3,"title":"Multiply the derivatives","content":"Now, we multiply the derivative of the outer function by the derivative of the inner function:\n\n$$\\frac{d}{dx}(x^2 + 10)^3 = 3(x^2 + 10)^2 \\times (2x)$$\n\nApplying this step correctly earns you the first method mark **(M1)**."},{"type":"text","order":4,"title":"Simplify the final answer","content":"Simplify the expression by multiplying the coefficients:\n\n$$3 \\times 2x = 6x$$\n\nThus, the final derivative is:\n\n$$6x(x^2 + 10)^2$$\n\nThis completes the calculation and earns you the final accuracy mark **(A1)**."}]},{"id":"a0081eec-278f-408c-bd4a-0ea5652e7338:algebraic_expansion","label":"(b) Expansion and Term-by-Term Differentiation","steps":[{"type":"text","order":0,"title":"Understand the method","content":"To differentiate the function $y = (x^2 + 10)^3$ using the **Expansion and Term-by-Term Differentiation** method, we will first fully expand the cubic expression algebraically. Once we have a polynomial in terms of powers of $x$, we can differentiate each term individually using the power rule:\n\n$$\\frac{d}{dx}(x^n) = nx^{n-1}$$"},{"type":"text","order":1,"title":"Expand the expression","content":"We can expand the expression $(x^2 + 10)^3$ using the binomial expansion formula for a cubic expression:\n\n$$(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$$\n\nSubstituting $a = x^2$ and $b = 10$, we get:\n\n$$\\begin{aligned} (x^2 + 10)^3 &= (x^2)^3 + 3(x^2)^2(10) + 3(x^2)(10)^2 + 10^3 \\\\ &= x^6 + 3(x^4)(10) + 3(x^2)(100) + 1000 \\\\ &= x^6 + 30x^4 + 300x^2 + 1000 \\end{aligned}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y = \\htmlData{anchor=term1}{\\color{@sky}{x^6}} + \\htmlData{anchor=term2}{\\color{@amber}{30x^4}} + \\htmlData{anchor=term3}{\\color{@purple}{300x^2}} + \\htmlData{anchor=term4}{\\color{@teal}{1000}}$"},{"latex":"$$\\frac{dy}{dx} = \\htmlData{anchor=diff1}{\\color{@sky}{6x^5}} + \\htmlData{anchor=diff2}{\\color{@amber}{120x^3}} + \\htmlData{anchor=diff3}{\\color{@purple}{600x}} + \\htmlData{anchor=diff4}{\\color{@teal}{0}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"term1"},{"color":"amber","style":"text-color","anchor":"term2"},{"color":"purple","style":"text-color","anchor":"term3"},{"color":"teal","style":"text-color","anchor":"term4"}],"connections":[{"to":"diff1","from":"term1","color":"sky","label":"Power Rule","style":"solid"},{"to":"diff2","from":"term2","color":"amber","label":"Power Rule","style":"solid"},{"to":"diff3","from":"term3","color":"purple","label":"Power Rule","style":"solid"},{"to":"diff4","from":"term4","color":"teal","label":"Constant Rule","style":"solid"}]},"order":2,"title":"Differentiate term-by-term","description":"Now, we apply the power rule to differentiate each term of the expanded polynomial with respect to $x$ individually."},{"type":"text","order":3,"title":"Factor the final answer","content":"To simplify our derivative and match it with the standard factored form, we can factor out the greatest common factor, which is $6x$:\n\n$$\\begin{aligned} \\frac{dy}{dx} &= 6x^5 + 120x^3 + 600x \\\\ &= 6x(x^4 + 20x^2 + 100) \\end{aligned}$$\n\nNotice that the quadratic expression inside the brackets is a perfect square trinomial:\n\n$$x^4 + 20x^2 + 100 = (x^2 + 10)^2$$\n\nSo, the fully factored derivative is:\n\n$$\\frac{dy}{dx} = 6x(x^2 + 10)^2$$"}]}],"generatedAt":"2026-06-09T02:59:20.990Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null}],"subjectCode":"a_level_maths__wjec","summaries":[{"id":"43670804-b308-4c43-b1b0-15ba09240572","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"6dd38f5a-9457-41f6-bdd8-5f51fb66d919","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"e0b9b73d-7217-4a30-9f07-a62b43ccabce","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"218d405b-a791-47c6-a02e-ddde693db591","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"233cef29-b34e-452d-b329-eeb2f8d1af50","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"423b5f9a-ebf5-4f40-82a1-09c4061ade2b","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":false},{"id":"5e22ee31-8d84-465e-9ae7-5f3393f1bbcc","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"6b16a43e-76ff-4df7-942f-0b53c44f143b","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":false},{"id":"8423de37-6e55-4383-8387-71cd38a57705","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"aabe4c72-d7d0-43ee-9c1d-1a1b6470204a","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":false},{"id":"d6515a96-3299-4394-97bc-a7029aa939de","difficultyBand":"easy","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"d7542af0-c3ba-44ed-81cd-ab6ab6f8a30a","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"dcc84617-b6eb-4564-b8e5-ed5142a90f67","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"e10def56-640b-416a-b974-27935fa57ad3","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"ed0d3408-c4dc-41f2-9018-0ac8079562aa","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"ee8cc8d3-1b89-4d8b-8908-fe61c14cee60","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"ef7406ef-b962-485d-967b-263dbf039539","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"f2d4c50a-dfb8-4a87-a753-18a9f23a84cc","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"fcc4935a-6dc7-4c9c-afb0-ba70d84f23a6","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":false},{"id":"66f47b42-061b-447f-af98-b82c19ee318b","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"783d7795-137b-4822-be83-d9c1d55fe34c","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":false},{"id":"a1ddd745-946c-48b3-92c5-c597b48104e1","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"c3f9a0ad-9898-4b93-8e8f-6da9a3e5e6b5","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"0720e9a2-40c4-4527-8636-f1396bc14a09","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"1fe24200-387c-4d5a-aadd-796df3778a45","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"324c01fc-3c23-4838-a251-2b61cef38926","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":false},{"id":"3982979a-c184-4cab-bb0c-9106d55688fd","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"4af8e31a-8a63-44a3-8769-01cd5abb9467","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"6c577d2b-91b1-48d8-b0d0-44c32c8af6a3","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"8aa09984-dd36-4279-bf61-a0159bfd11d4","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":false},{"id":"9548d2cd-09f8-45c9-8239-a49aa6d6bb6d","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"9ec7b6f4-b03d-4589-a8f9-3cce13a23630","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"a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Differentiation (A-level only)","topicId":"929a6783-d6e8-4c97-ae6b-5d91a25392ec","topicSidebar":"$L3c","topicSlug":"differentiation-13","topicTitle":"3.6.5 Differentiation (A-level only)"}]}],"$L3d","$L3e"]}]

3.6.5 Differentiation (A-level only)

Question 3

3.6.5 Differentiation (A-level only) Questions

3.6.5 Differentiation (A-level only)

Question 3

3.6.5 Differentiation (A-level only) Questions