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below:\n\n\$\\frac{d\\theta}{dt} = -9e^{-0.04t}\$\n\n\$\\frac{d\\theta}{dt} = 9e^{-0.04t}\$\n\n\$\\frac{d\\theta}{dt} = -5625e^{-0.04t}\$\n\n\$\\frac{d\\theta}{dt} = -0.04e^{-0.04t}\$","markscheme":"- Realises that the rate of change is the derivative of temperature with respect to time $\\frac{d\\theta}{dt}$ M1\n- Differentiates using the rule $\\frac{d}{dt}(e^{kt}) = ke^{kt}$ to obtain $225 \\times (-0.04)e^{-0.04t}$ M1\n- Calculates $225 \\times (-0.04) = -9$ to give the expression $-9e^{-0.04t}$ A1","marks":3,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"direct_differentiation","label":"Direct Differentiation","steps":[{"type":"text","order":0,"title":"Identify the goal","content":"We want to find the rate of change of the coolant's temperature, which is mathematically represented by the derivative of the temperature with respect to time, $\\frac{d\\theta}{dt}$.\n\nThe equation given is:\n\n$$\\theta = 225e^{-0.04t}$$"},{"type":"text","order":1,"title":"Recall the derivative rule","content":"To differentiate an exponential term of the form $e^{kt}$ with respect to $t$, we use the standard exponential differentiation rule:\n\n$$\\frac{d}{dt}(e^{kt}) = k e^{kt}$$\n\nIf we have a constant coefficient in front, we simply multiply the constant by the derivative of the exponential part."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\theta = 225e^{\\htmlData{anchor=k-val}{\\color{@amber}{-0.04}}t}$"},{"latex":"$$\\frac{d\\theta}{dt} = 225 \\times \\htmlData{anchor=k-deriv}{\\color{@amber}{(-0.04)}}e^{\\htmlData{anchor=k-exp}{\\color{@amber}{-0.04}}t}$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"k-val"},{"color":"amber","style":"text-color","anchor":"k-deriv"}],"connections":[{"to":"k-deriv","from":"k-val","color":"amber","label":"multiply by k","style":"solid"}]},"order":2,"title":"Differentiate the function","description":"Applying the rule where k = -0.04, we pull the factor of -0.04 out to multiply by the constant 225."},{"type":"text","order":3,"title":"Calculate the coefficient","content":"Now, multiply the constant coefficient by the exponent's coefficient:\n\n$$225 \\times (-0.04) = -9$$\n\nSubstituting this back gives the final rate of change expression:\n\n$$\\frac{d\\theta}{dt} = -9e^{-0.04t}$$"}]},{"id":"chain_rule_substitution","label":"Chain Rule Substitution","steps":[{"type":"text","order":0,"title":"Identify the goal","content":"We need to find an expression for the rate of change of the coolant's temperature with respect to time, which is the derivative $\\frac{d\\theta}{dt}$.\n\nThe given temperature equation is:\n\n$$\\theta = 225e^{-0.04t}$$"},{"type":"hyper_latex","block":{"nodes":[{"text":"Using the Chain Rule, we can define a simpler intermediate variable u to represent the exponent:"},{"latex":"$$\\htmlData{anchor=u-def}{\\color{@sky}{u}} = \\htmlData{anchor=exponent}{\\color{@sky}{-0.04t}}$"},{"text":"This allows us to write the temperature equation in terms of u:"},{"latex":"$$\\htmlData{anchor=theta-eq}{\\theta} = 225e^{\\htmlData{anchor=u-sub}{\\color{@sky}{u}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-def"},{"color":"sky","style":"text-color","anchor":"exponent"},{"color":"sky","style":"text-color","anchor":"u-sub"}],"connections":[{"to":"u-sub","from":"u-def","color":"sky","label":"substitute","style":"solid"}]},"order":1,"title":"Define intermediate variable","description":"We substitute u = -0.04t to simplify the expression into a standard exponential form."},{"type":"hyper_latex","block":{"nodes":[{"text":"Next, we differentiate both equations with respect to their corresponding variables."},{"text":"First, differentiate u with respect to t:"},{"latex":"$$\\htmlData{anchor=dudt-eq}{\\frac{du}{dt}} = \\htmlData{anchor=dudt-val}{\\color{@amber}{-0.04}}$"},{"text":"Then, differentiate theta with respect to u:"},{"latex":"$$\\htmlData{anchor=dthdu-eq}{\\frac{d\\theta}{du}} = \\htmlData{anchor=dthdu-val}{\\color{@pink}{225e^u}}$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"dudt-val"},{"color":"pink","style":"text-color","anchor":"dthdu-val"}],"connections":[]},"order":2,"title":"Find individual derivatives","description":"We calculate the derivatives with respect to t and u to prepare for the Chain Rule."},{"type":"hyper_latex","block":{"nodes":[{"text":"The Chain Rule states:"},{"latex":"$$\\frac{d\\theta}{dt} = \\htmlData{anchor=dthdu-link}{\\color{@pink}{\\frac{d\\theta}{du}}} \\times \\htmlData{anchor=dudt-link}{\\color{@amber}{\\frac{du}{dt}}}$"},{"text":"Substituting our derivatives into the formula:"},{"latex":"$$\\frac{d\\theta}{dt} = (\\htmlData{anchor=dthdu-sub}{\\color{@pink}{225e^u}}) \\times (\\htmlData{anchor=dudt-sub}{\\color{@amber}{-0.04}})$"}],"highlights":[{"color":"pink","style":"text-color","anchor":"dthdu-link"},{"color":"amber","style":"text-color","anchor":"dudt-link"},{"color":"pink","style":"text-color","anchor":"dthdu-sub"},{"color":"amber","style":"text-color","anchor":"dudt-sub"}],"connections":[{"to":"dthdu-sub","from":"dthdu-link","color":"pink","label":null,"style":"dashed"},{"to":"dudt-sub","from":"dudt-link","color":"amber","label":null,"style":"dashed"}]},"order":3,"title":"Apply Chain Rule","description":"Using the Chain Rule formula, we multiply the two intermediate derivatives together."},{"type":"text","order":4,"title":"Simplify and substitute back","content":"Now, we multiply the constants and substitute our expression for $u$ back into the formula:\n\n$$\\begin{aligned} \\frac{d\\theta}{dt} &= 225 \\times (-0.04)e^u \\\\ &= -9e^u \\end{aligned}$$\n\nSubstituting $u = -0.04t$ back in gives:\n\n$$\\frac{d\\theta}{dt} = -9e^{-0.04t}$$"}]}],"generatedAt":"2026-06-09T04:48:15.479Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"545718a2-d856-48b5-ba3b-1f24c96bcd75:direct_differentiation","label":"Direct Differentiation","steps":[{"type":"text","order":0,"title":"Identify the goal","content":"We want to find the rate of change of the coolant's temperature, which is mathematically represented by the derivative of the temperature with respect to time, $\\frac{d\\theta}{dt}$.\n\nThe equation given is:\n\n$$\\theta = 225e^{-0.04t}$$"},{"type":"text","order":1,"title":"Recall the derivative rule","content":"To differentiate an exponential term of the form $e^{kt}$ with respect to $t$, we use the standard exponential differentiation rule:\n\n$$\\frac{d}{dt}(e^{kt}) = k e^{kt}$$\n\nIf we have a constant coefficient in front, we simply multiply the constant by the derivative of the exponential part."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\theta = 225e^{\\htmlData{anchor=k-val}{\\color{@amber}{-0.04}}t}$"},{"latex":"$$\\frac{d\\theta}{dt} = 225 \\times \\htmlData{anchor=k-deriv}{\\color{@amber}{(-0.04)}}e^{\\htmlData{anchor=k-exp}{\\color{@amber}{-0.04}}t}$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"k-val"},{"color":"amber","style":"text-color","anchor":"k-deriv"}],"connections":[{"to":"k-deriv","from":"k-val","color":"amber","label":"multiply by k","style":"solid"}]},"order":2,"title":"Differentiate the function","description":"Applying the rule where k = -0.04, we pull the factor of -0.04 out to multiply by the constant 225."},{"type":"text","order":3,"title":"Calculate the coefficient","content":"Now, multiply the constant coefficient by the exponent's coefficient:\n\n$$225 \\times (-0.04) = -9$$\n\nSubstituting this back gives the final rate of change expression:\n\n$$\\frac{d\\theta}{dt} = -9e^{-0.04t}$$"}]},{"id":"545718a2-d856-48b5-ba3b-1f24c96bcd75:chain_rule_substitution","label":"Chain Rule Substitution","steps":[{"type":"text","order":0,"title":"Identify the goal","content":"We need to find an expression for the rate of change of the coolant's temperature with respect to time, which is the derivative $\\frac{d\\theta}{dt}$.\n\nThe given temperature equation is:\n\n$$\\theta = 225e^{-0.04t}$$"},{"type":"hyper_latex","block":{"nodes":[{"text":"Using the Chain Rule, we can define a simpler intermediate variable u to represent the exponent:"},{"latex":"$$\\htmlData{anchor=u-def}{\\color{@sky}{u}} = \\htmlData{anchor=exponent}{\\color{@sky}{-0.04t}}$"},{"text":"This allows us to write the temperature equation in terms of u:"},{"latex":"$$\\htmlData{anchor=\\theta-eq}{\\theta} = 225e^{\\htmlData{anchor=u-sub}{\\color{@sky}{u}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-def"},{"color":"sky","style":"text-color","anchor":"exponent"},{"color":"sky","style":"text-color","anchor":"u-sub"}],"connections":[{"to":"u-sub","from":"u-def","color":"sky","label":"substitute","style":"solid"}]},"order":1,"title":"Define intermediate variable","description":"We substitute u = -0.04t to simplify the expression into a standard exponential form."},{"type":"hyper_latex","block":{"nodes":[{"text":"Next, we differentiate both equations with respect to their corresponding variables."},{"text":"First, differentiate u with respect to t:"},{"latex":"$$\\htmlData{anchor=dudt-eq}{\\frac{du}{dt}} = \\htmlData{anchor=dudt-val}{\\color{@amber}{-0.04}}$"},{"text":"Then, differentiate theta with respect to u:"},{"latex":"$$\\htmlData{anchor=dthdu-eq}{\\frac{d\\theta}{du}} = \\htmlData{anchor=dthdu-val}{\\color{@pink}{225e^u}}$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"dudt-val"},{"color":"pink","style":"text-color","anchor":"dthdu-val"}],"connections":[]},"order":2,"title":"Find individual derivatives","description":"We calculate the derivatives with respect to t and u to prepare for the Chain Rule."},{"type":"hyper_latex","block":{"nodes":[{"text":"The Chain Rule states:"},{"latex":"$$\\frac{d\\theta}{dt} = \\htmlData{anchor=dthdu-link}{\\color{@pink}{\\frac{d\\theta}{du}}} \\times \\htmlData{anchor=dudt-link}{\\color{@amber}{\\frac{du}{dt}}}$"},{"text":"Substituting our derivatives into the formula:"},{"latex":"$$\\frac{d\\theta}{dt} = (\\htmlData{anchor=dthdu-sub}{\\color{@pink}{225e^u}}) \\times (\\htmlData{anchor=dudt-sub}{\\color{@amber}{-0.04}})$"}],"highlights":[{"color":"pink","style":"text-color","anchor":"dthdu-link"},{"color":"amber","style":"text-color","anchor":"dudt-link"},{"color":"pink","style":"text-color","anchor":"dthdu-sub"},{"color":"amber","style":"text-color","anchor":"dudt-sub"}],"connections":[{"to":"dthdu-sub","from":"dthdu-link","color":"pink","label":null,"style":"dashed"},{"to":"dudt-sub","from":"dudt-link","color":"amber","label":null,"style":"dashed"}]},"order":3,"title":"Apply Chain Rule","description":"Using the Chain Rule formula, we multiply the two intermediate derivatives together."},{"type":"text","order":4,"title":"Simplify and substitute back","content":"Now, we multiply the constants and substitute our expression for $u$ back into the formula:\n\n$$\\begin{aligned} \\frac{d\\theta}{dt} &= 225 \\times (-0.04)e^u \\\\ &= -9e^u \\end{aligned}$$\n\nSubstituting $u = -0.04t$ back in gives:\n\n$$\\frac{d\\theta}{dt} = -9e^{-0.04t}$$"}]}],"generatedAt":"2026-06-09T04:48:40.284Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"8081f0f9-1174-4cd4-abe3-673dc009566a","revisionId":"97f307a5-3cd9-43db-ab6a-0f230400847f","questionNumber":null,"questionType":"LA","specification":"A curve has the equation \$y = x^5 + 3^x\$.","difficulty":3,"difficultyBand":"easy","tier":null,"totalMarks":2,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"f2b18411-7941-4e19-b63b-ddbfcb31cca6","label":null,"content":"Find an expression for \$\\dfrac{dy}{dx}\$.","markscheme":"- Differentiates $x^5$ to obtain $5x^4$ B1\n- Differentiates $3^x$ to obtain $3^x \\ln 3$ B1","marks":2,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"direct_rules","label":"Direct Differentiation Rules","steps":[{"type":"text","order":0,"title":"Identify the terms","content":"To find $\\dfrac{dy}{dx}$ for the curve $y = x^5 + 3^x$, we can differentiate each term separately using standard differentiation rules.\n\nThe equation has two terms:\n1. A power term: $x^5$\n2. An exponential term with base 3: $3^x$","highlights":[{"text":"y = x^5 + 3^x","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$\\frac{d}{dx}\\left(\\htmlData{anchor=power-term}{\\color{@sky}{x^n}}\\right) = \\htmlData{anchor=power-rule}{\\color{@sky}{n x^{n-1}}}$$"},{"latex":"$$$\\frac{d}{dx}\\left(\\htmlData{anchor=x5-term}{\\color{@sky}{x^5}}\\right) = \\htmlData{anchor=x5-diff}{\\color{@sky}{5x^4}}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"power-term"},{"color":"sky","style":"text-color","anchor":"x5-term"}],"connections":[{"to":"x5-term","from":"power-term","color":"sky","label":"substitute n=5","style":"solid"},{"to":"x5-diff","from":"power-rule","color":"sky","label":"power rule result","style":"solid"}]},"order":1,"title":"Differentiate the power term","description":"We apply the power rule $\\dfrac{d}{dx}(x^n) = n x^{n-1}$ to the first term, where $n = 5$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$\\frac{d}{dx}\\left(\\htmlData{anchor=exp-term}{\\color{@purple}{a^x}}\\right) = \\htmlData{anchor=exp-rule}{\\color{@purple}{a^x \\ln a}}$$"},{"latex":"$$$\\frac{d}{dx}\\left(\\htmlData{anchor=3x-term}{\\color{@purple}{3^x}}\\right) = \\htmlData{anchor=3x-diff}{\\color{@purple}{3^x \\ln 3}}$$"}],"highlights":[{"color":"purple","style":"text-color","anchor":"exp-term"},{"color":"purple","style":"text-color","anchor":"3x-term"}],"connections":[{"to":"3x-term","from":"exp-term","color":"purple","label":"substitute a=3","style":"solid"},{"to":"3x-diff","from":"exp-rule","color":"purple","label":"exponential rule result","style":"solid"}]},"order":2,"title":"Differentiate the exponential term","description":"We apply the general exponential derivative rule $\\dfrac{d}{dx}(a^x) = a^x \\ln a$ to the second term, where $a = 3$."},{"type":"text","order":3,"title":"Combine the results","content":"Now, we combine the derivatives of each individual term to get our final expression for $\\dfrac{dy}{dx}$:\n\n$$\\frac{dy}{dx} = 5x^4 + 3^x \\ln 3$$\n\n### Markscheme Breakdown\n- **1 mark (B1)**: Correctly differentiating $x^5$ to obtain $5x^4$\n- **1 mark (B1)**: Correctly differentiating $3^x$ to obtain $3^x \\ln 3$"}]},{"id":"base_e_conversion","label":"Converting to Base e","steps":[{"type":"text","order":0,"title":"Analyze the equation","content":"We need to find the derivative of the curve given by:\n\n$$y = x^5 + 3^x$$\n\nTo differentiate this function, we can differentiate each term separately using the sum rule:\n\n$$\\frac{dy}{dx} = \\frac{d}{dx}\\left(x^5\\right) + \\frac{d}{dx}\\left(3^x\\right)$$"},{"type":"text","order":1,"title":"Differentiate the first term","content":"For the first term, we apply the power rule of differentiation, which states that:\n\n$$\\frac{d}{dx}\\left(x^n\\right) = nx^{n-1}$$\n\nApplying this with $n = 5$ gives:\n\n$$\\frac{d}{dx}\\left(x^5\\right) = 5x^4$$\n\n*(This earns the first mark, B1)*","highlights":[{"text":"Differentiates $x^5$ to obtain $5x^4$","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$3^x = \\htmlData{anchor=base-e}{\\color{@sky}{e^{\\ln(3^x)}}}$$"},{"latex":"$$$3^x = \\htmlData{anchor=rule-apply}{\\color{@sky}{e^{x \\ln 3}}}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"base-e"},{"color":"sky","style":"text-color","anchor":"rule-apply"}],"connections":[{"to":"rule-apply","from":"base-e","color":"sky","label":"power rule of logs","style":"solid"}]},"order":2,"title":"Convert to Base e","description":"To differentiate the exponential term $3^x$, we rewrite it using the natural base $e$. Since exponential and logarithmic functions are inverses, we can write $3 = e^{\\ln 3}$, which means:"},{"type":"text","order":3,"title":"Differentiate using Chain Rule","content":"Now we differentiate $e^{x \\ln 3}$ with respect to $x$ using the chain rule.\n\nRecall that:\n\n$$\\frac{d}{dx}\\left(e^{kx}\\right) = k e^{kx}$$\n\nHere, the constant is $k = \\ln 3$. Applying this rule:\n\n$$\\frac{d}{dx}\\left(e^{x \\ln 3}\\right) = (\\ln 3)e^{x \\ln 3}$$\n\nSince we defined $e^{x \\ln 3} = 3^x$, we can substitute it back:\n\n$$\\frac{d}{dx}\\left(3^x\\right) = 3^x \\ln 3$$\n\n*(This earns the second mark, B1)*","highlights":[{"text":"Differentiates $3^x$ to obtain $3^x \\ln 3$","location":"specification"}]},{"type":"text","order":4,"title":"Combine the results","content":"Finally, we combine both derivative terms to obtain the overall derivative:\n\n$$\\frac{dy}{dx} = 5x^4 + 3^x \\ln 3$$"}]}],"generatedAt":"2026-06-09T06:11:54.275Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"f2b18411-7941-4e19-b63b-ddbfcb31cca6:direct_rules","label":"Direct Differentiation Rules","steps":[{"type":"text","order":0,"title":"Identify the terms","content":"To find $\\dfrac{dy}{dx}$ for the curve $y = x^5 + 3^x$, we can differentiate each term separately using standard differentiation rules.\n\nThe equation has two terms:\n1. A power term: $x^5$\n2. An exponential term with base 3: $3^x$","highlights":[{"text":"y = x^5 + 3^x","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$\\frac{d}{dx}\\left(\\htmlData{anchor=power-term}{\\color{@sky}{x^n}}\\right) = \\htmlData{anchor=power-rule}{\\color{@sky}{n x^{n-1}}}$$"},{"latex":"$$$\\frac{d}{dx}\\left(\\htmlData{anchor=x5-term}{\\color{@sky}{x^5}}\\right) = \\htmlData{anchor=x5-diff}{\\color{@sky}{5x^4}}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"power-term"},{"color":"sky","style":"text-color","anchor":"x5-term"}],"connections":[{"to":"x5-term","from":"power-term","color":"sky","label":"substitute n=5","style":"solid"},{"to":"x5-diff","from":"power-rule","color":"sky","label":"power rule result","style":"solid"}]},"order":1,"title":"Differentiate the power term","description":"We apply the power rule $\\dfrac{d}{dx}(x^n) = n x^{n-1}$ to the first term, where $n = 5$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$\\frac{d}{dx}\\left(\\htmlData{anchor=exp-term}{\\color{@purple}{a^x}}\\right) = \\htmlData{anchor=exp-rule}{\\color{@purple}{a^x \\ln a}}$$"},{"latex":"$$$\\frac{d}{dx}\\left(\\htmlData{anchor=3x-term}{\\color{@purple}{3^x}}\\right) = \\htmlData{anchor=3x-diff}{\\color{@purple}{3^x \\ln 3}}$$"}],"highlights":[{"color":"purple","style":"text-color","anchor":"exp-term"},{"color":"purple","style":"text-color","anchor":"3x-term"}],"connections":[{"to":"3x-term","from":"exp-term","color":"purple","label":"substitute a=3","style":"solid"},{"to":"3x-diff","from":"exp-rule","color":"purple","label":"exponential rule result","style":"solid"}]},"order":2,"title":"Differentiate the exponential term","description":"We apply the general exponential derivative rule $\\dfrac{d}{dx}(a^x) = a^x \\ln a$ to the second term, where $a = 3$."},{"type":"text","order":3,"title":"Combine the results","content":"Now, we combine the derivatives of each individual term to get our final expression for $\\dfrac{dy}{dx}$:\n\n$$\\frac{dy}{dx} = 5x^4 + 3^x \\ln 3$$\n\n### Markscheme Breakdown\n- **1 mark (B1)**: Correctly differentiating $x^5$ to obtain $5x^4$\n- **1 mark (B1)**: Correctly differentiating $3^x$ to obtain $3^x \\ln 3$"}]},{"id":"f2b18411-7941-4e19-b63b-ddbfcb31cca6:base_e_conversion","label":"Converting to Base e","steps":[{"type":"text","order":0,"title":"Analyze the equation","content":"We need to find the derivative of the curve given by:\n\n$$y = x^5 + 3^x$$\n\nTo differentiate this function, we can differentiate each term separately using the sum rule:\n\n$$\\frac{dy}{dx} = \\frac{d}{dx}\\left(x^5\\right) + \\frac{d}{dx}\\left(3^x\\right)$$"},{"type":"text","order":1,"title":"Differentiate the first term","content":"For the first term, we apply the power rule of differentiation, which states that:\n\n$$\\frac{d}{dx}\\left(x^n\\right) = nx^{n-1}$$\n\nApplying this with $n = 5$ gives:\n\n$$\\frac{d}{dx}\\left(x^5\\right) = 5x^4$$\n\n*(This earns the first mark, B1)*","highlights":[{"text":"Differentiates $x^5$ to obtain $5x^4$","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$3^x = \\htmlData{anchor=base-e}{\\color{@sky}{e^{\\ln(3^x)}}}$$"},{"latex":"$$$3^x = \\htmlData{anchor=rule-apply}{\\color{@sky}{e^{x \\ln 3}}}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"base-e"},{"color":"sky","style":"text-color","anchor":"rule-apply"}],"connections":[{"to":"rule-apply","from":"base-e","color":"sky","label":"power rule of logs","style":"solid"}]},"order":2,"title":"Convert to Base e","description":"To differentiate the exponential term $3^x$, we rewrite it using the natural base $e$. Since exponential and logarithmic functions are inverses, we can write $3 = e^{\\ln 3}$, which means:"},{"type":"text","order":3,"title":"Differentiate using Chain Rule","content":"Now we differentiate $e^{x \\ln 3}$ with respect to $x$ using the chain rule.\n\nRecall that:\n\n$$\\frac{d}{dx}\\left(e^{kx}\\right) = k e^{kx}$$\n\nHere, the constant is $k = \\ln 3$. Applying this rule:\n\n$$\\frac{d}{dx}\\left(e^{x \\ln 3}\\right) = (\\ln 3)e^{x \\ln 3}$$\n\nSince we defined $e^{x \\ln 3} = 3^x$, we can substitute it back:\n\n$$\\frac{d}{dx}\\left(3^x\\right) = 3^x \\ln 3$$\n\n*(This earns the second mark, B1)*","highlights":[{"text":"Differentiates $3^x$ to obtain $3^x \\ln 3$","location":"specification"}]},{"type":"text","order":4,"title":"Combine the results","content":"Finally, we combine both derivative terms to obtain the overall derivative:\n\n$$\\frac{dy}{dx} = 5x^4 + 3^x \\ln 3$$"}]}],"generatedAt":"2026-06-09T06:12:04.456Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"9768deb2-67df-4442-88c8-1ff2002600a0","revisionId":"2ecb217a-a23b-4208-ae1d-944a2fa61ab8","questionNumber":null,"questionType":"LA","specification":"","difficulty":3,"difficultyBand":"easy","tier":null,"totalMarks":2,"sourceTitle":null,"calculatorUse":null,"isCalculation":false,"parts":[{"id":"3b1756ba-56ab-48f5-80de-f5d267d421dd","label":null,"content":"Given that \$y = \\ln(7x)\$, find \$\\frac{dy}{dx}\$. Select the correct option:\n\nA: \$\\frac{dy}{dx} = \\frac{1}{x}\$\n\nB: \$\\frac{dy}{dx} = \\frac{7}{x}\$\n\nC: \$\\frac{dy}{dx} = \\frac{1}{7x}\$\n\nD: \$\\frac{dy}{dx} = \\ln(7)\$","markscheme":"- Uses the chain rule with $u = 7x$ to find $\\frac{dy}{du} = \\frac{1}{u} = \\frac{1}{7x}$ and $\\frac{du}{dx} = 7$, or uses laws of logarithms to rewrite $y = \\ln(7) + \\ln(x)$ M1\n- Multiplies the chain rule components $\\frac{1}{7x} \\times 7$ or differentiates the split log terms to obtain $\\frac{dy}{dx} = \\frac{1}{x}$ A1","marks":2,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"chain_rule","label":"Using Chain Rule","steps":[{"type":"text","order":0,"title":"Identify the composite function","content":"The given function is $y = \\ln(7x)$. This is a composite function, meaning it has an \"inner function\" nested inside an \"outer function\". We can differentiate this using the **chain rule**.","highlights":[{"text":"y = \\ln(7x)","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"text":"Let us define the inner function as $u$:"},{"latex":"$$\\htmlData{anchor=u-def}{\\color{@sky}{u}} = \\htmlData{anchor=u-expr}{\\color{@sky}{7x}}$"},{"text":"This simplifies our original function to:"},{"latex":"$$y = \\ln(\\htmlData{anchor=u-sub}{\\color{@sky}{u}})$"}],"highlights":[{"color":"sky","style":"box","anchor":"u-def"},{"color":"sky","style":"box","anchor":"u-sub"}],"connections":[{"to":"u-sub","from":"u-def","color":"sky","label":"substitute u","style":"solid"}]},"order":1,"title":"Define the substitution","description":"We substitute the inner function $7x$ with a new variable $u$ to make it easier to differentiate."},{"type":"text","order":2,"title":"Find the individual derivatives","content":"Now we find the derivative of both parts:\n\n1. Differentiate $y = \\ln(u)$ with respect to $u$:\n$$\\frac{dy}{du} = \\frac{1}{u}$$\n\n2. Differentiate $u = 7x$ with respect to $x$:\n$$\\frac{du}{dx} = 7$$"},{"type":"hyper_latex","block":{"nodes":[{"text":"The chain rule states:"},{"latex":"$$\\frac{dy}{dx} = \\htmlData{anchor=dy-du}{\\color{@pink}{\\frac{dy}{du}}} \\times \\htmlData{anchor=du-dx}{\\color{@purple}{\\frac{du}{dx}}}$"},{"text":"Substituting our derivatives in:"},{"latex":"$$\\frac{dy}{dx} = \\htmlData{anchor=dy-du-sub}{\\color{@pink}{\\frac{1}{u}}} \\times \\htmlData{anchor=du-dx-sub}{\\color{@purple}{7}}$"}],"highlights":[{"color":"pink","style":"text-color","anchor":"dy-du"},{"color":"purple","style":"text-color","anchor":"du-dx"}],"connections":[{"to":"dy-du-sub","from":"dy-du","color":"pink","label":null,"style":"solid"},{"to":"du-dx-sub","from":"du-dx","color":"purple","label":null,"style":"solid"}]},"order":3,"title":"Apply the Chain Rule","description":"We multiply the two derivatives together as defined by the chain rule."},{"type":"text","order":4,"title":"Substitute back and simplify","content":"Now substitute $u = 7x$ back into the expression:\n\n$$\\frac{dy}{dx} = \\frac{1}{7x} \\times 7$$\n\nSimplifying the expression by cancelling the factor of $7$ in the numerator and denominator:\n\n$$\\frac{dy}{dx} = \\frac{1}{x}$$"},{"type":"text","order":5,"title":"Select the correct option","content":"Comparing our result with the options provided:\n\n* **A**: $\\frac{dy}{dx} = \\frac{1}{x}$\n* **B**: $\\frac{dy}{dx} = \\frac{7}{x}$\n* **C**: $\\frac{dy}{dx} = \\frac{1}{7x}$\n* **D**: $\\frac{dy}{dx} = \\ln(7)$\n\nThe correct option is **A**.","highlights":[{"text":"A: \$\\frac{dy}{dx} = \\frac{1}{x}\$","index":0,"location":"option"},{"text":"B: \$\\frac{dy}{dx} = \\frac{7}{x}\$","index":1,"location":"eliminate"},{"text":"C: \$\\frac{dy}{dx} = \\frac{1}{7x}\$","index":2,"location":"eliminate"},{"text":"D: \$\\frac{dy}{dx} = \\ln(7)\$","index":3,"location":"eliminate"}]}]},{"id":"logarithm_properties","label":"Using Logarithm Properties","steps":[{"type":"text","order":0,"title":"Recall logarithm properties","content":"To make differentiation straightforward, we can use the multiplication law of logarithms:\n\n$$\\ln(ab) = \\ln(a) + \\ln(b)$$\n\nApplying this property to our function, we can rewrite the expression:\n\n$$y = \\ln(7x) = \\ln(7) + \\ln(x)$$","highlights":[{"text":"y = \\ln(7x)","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y = \\htmlData{anchor=const}{\\ln(7)} + \\htmlData{anchor=var}{\\ln(x)}$"},{"latex":"$$\\frac{dy}{dx} = \\htmlData{anchor=const-diff}{0} + \\htmlData{anchor=var-diff}{\\frac{1}{x}}$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"const"},{"color":"sky","style":"text-color","anchor":"var"}],"connections":[{"to":"const-diff","from":"const","color":"amber","label":"constant","style":"solid"},{"to":"var-diff","from":"var","color":"sky","label":"derivative","style":"solid"}]},"order":1,"title":"Differentiate term-by-term","description":"Now, we differentiate each term with respect to $x$:\n\n1. Since $7$ is a constant, $\\ln(7)$ is also a constant value. The derivative of any constant is $0$.\n2. The derivative of $\\ln(x)$ with respect to $x$ is a standard derivative:\n\n$$\\frac{d}{dx}(\\ln(x)) = \\frac{1}{x}$$"},{"type":"text","order":2,"title":"Simplify and match","content":"Adding the terms together, we get:\n\n$$\\frac{dy}{dx} = \\frac{1}{x}$$\n\nThis matches **Option A**.","highlights":[{"text":"A: \\frac{dy}{dx} = \\frac{1}{x}","index":0,"location":"option"}]}]}],"generatedAt":"2026-06-09T07:23:54.904Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"3b1756ba-56ab-48f5-80de-f5d267d421dd:chain_rule","label":"Using Chain Rule","steps":[{"type":"text","order":0,"title":"Identify the composite function","content":"The given function is $y = \\ln(7x)$. This is a composite function, meaning it has an \"inner function\" nested inside an \"outer function\". We can differentiate this using the **chain rule**.","highlights":[{"text":"y = \\ln(7x)","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"text":"Let us define the inner function as $u$:"},{"latex":"$$\\htmlData{anchor=u-def}{\\color{@sky}{u}} = \\htmlData{anchor=u-expr}{\\color{@sky}{7x}}$"},{"text":"This simplifies our original function to:"},{"latex":"$$y = \\ln(\\htmlData{anchor=u-sub}{\\color{@sky}{u}})$"}],"highlights":[{"color":"sky","style":"box","anchor":"u-def"},{"color":"sky","style":"box","anchor":"u-sub"}],"connections":[{"to":"u-sub","from":"u-def","color":"sky","label":"substitute u","style":"solid"}]},"order":1,"title":"Define the substitution","description":"We substitute the inner function $7x$ with a new variable $u$ to make it easier to differentiate."},{"type":"text","order":2,"title":"Find the individual derivatives","content":"Now we find the derivative of both parts:\n\n1. Differentiate $y = \\ln(u)$ with respect to $u$:\n$$\\frac{dy}{du} = \\frac{1}{u}$$\n\n2. Differentiate $u = 7x$ with respect to $x$:\n$$\\frac{du}{dx} = 7$$"},{"type":"hyper_latex","block":{"nodes":[{"text":"The chain rule states:"},{"latex":"$$\\frac{dy}{dx} = \\htmlData{anchor=dy-du}{\\color{@pink}{\\frac{dy}{du}}} \\times \\htmlData{anchor=du-dx}{\\color{@purple}{\\frac{du}{dx}}}$"},{"text":"Substituting our derivatives in:"},{"latex":"$$\\frac{dy}{dx} = \\htmlData{anchor=dy-du-sub}{\\color{@pink}{\\frac{1}{u}}} \\times \\htmlData{anchor=du-dx-sub}{\\color{@purple}{7}}$"}],"highlights":[{"color":"pink","style":"text-color","anchor":"dy-du"},{"color":"purple","style":"text-color","anchor":"du-dx"}],"connections":[{"to":"dy-du-sub","from":"dy-du","color":"pink","label":null,"style":"solid"},{"to":"du-dx-sub","from":"du-dx","color":"purple","label":null,"style":"solid"}]},"order":3,"title":"Apply the Chain Rule","description":"We multiply the two derivatives together as defined by the chain rule."},{"type":"text","order":4,"title":"Substitute back and simplify","content":"Now substitute $u = 7x$ back into the expression:\n\n$$\\frac{dy}{dx} = \\frac{1}{7x} \\times 7$$\n\nSimplifying the expression by cancelling the factor of $7$ in the numerator and denominator:\n\n$$\\frac{dy}{dx} = \\frac{1}{x}$$"},{"type":"text","order":5,"title":"Select the correct option","content":"Comparing our result with the options provided:\n\n* **A**: $\\frac{dy}{dx} = \\frac{1}{x}$\n* **B**: $\\frac{dy}{dx} = \\frac{7}{x}$\n* **C**: $\\frac{dy}{dx} = \\frac{1}{7x}$\n* **D**: $\\frac{dy}{dx} = \\ln(7)$\n\nThe correct option is **A**.","highlights":[{"text":"A: \$\\frac{dy}{dx} = \\frac{1}{x}\$","index":0,"location":"option"},{"text":"B: \$\\frac{dy}{dx} = \\frac{7}{x}\$","index":1,"location":"eliminate"},{"text":"C: \$\\frac{dy}{dx} = \\frac{1}{7x}\$","index":2,"location":"eliminate"},{"text":"D: \$\\frac{dy}{dx} = \\ln(7)\$","index":3,"location":"eliminate"}]}]},{"id":"3b1756ba-56ab-48f5-80de-f5d267d421dd:logarithm_properties","label":"Using Logarithm Properties","steps":[{"type":"text","order":0,"title":"Recall logarithm properties","content":"To make differentiation straightforward, we can use the multiplication law of logarithms:\n\n$$\\ln(ab) = \\ln(a) + \\ln(b)$$\n\nApplying this property to our function, we can rewrite the expression:\n\n$$y = \\ln(7x) = \\ln(7) + \\ln(x)$$","highlights":[{"text":"y = \\ln(7x)","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y = \\htmlData{anchor=const}{\\ln(7)} + \\htmlData{anchor=var}{\\ln(x)}$"},{"latex":"$$\\frac{dy}{dx} = \\htmlData{anchor=const-diff}{0} + \\htmlData{anchor=var-diff}{\\frac{1}{x}}$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"const"},{"color":"sky","style":"text-color","anchor":"var"}],"connections":[{"to":"const-diff","from":"const","color":"amber","label":"constant","style":"solid"},{"to":"var-diff","from":"var","color":"sky","label":"derivative","style":"solid"}]},"order":1,"title":"Differentiate term-by-term","description":"Now, we differentiate each term with respect to $x$:\n\n1. Since $7$ is a constant, $\\ln(7)$ is also a constant value. The derivative of any constant is $0$.\n2. The derivative of $\\ln(x)$ with respect to $x$ is a standard derivative:\n\n$$\\frac{d}{dx}(\\ln(x)) = \\frac{1}{x}$$"},{"type":"text","order":2,"title":"Simplify and match","content":"Adding the terms together, we get:\n\n$$\\frac{dy}{dx} = \\frac{1}{x}$$\n\nThis matches **Option A**.","highlights":[{"text":"A: \\frac{dy}{dx} = \\frac{1}{x}","index":0,"location":"option"}]}]}],"generatedAt":"2026-06-09T07:24:13.319Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null}],"subjectCode":"a_level_maths__ocr_mei","summaries":[{"id":"033c36da-c44c-4d13-8fc0-e3dac3b717ed","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":false},{"id":"969d301a-0b61-49f8-8434-489befd77816","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"638c9f27-a5eb-42f1-a612-72ee010d7ca3","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":false},{"id":"67e5705a-9017-4878-829d-8d83496303ea","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"7dc21b04-229b-488a-bb44-bdcb4b9571cc","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"8c1b5cbd-8fa6-45f7-bffe-63b0d4caff7e","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"ab0cabd7-bab3-4117-ac60-72945729dbac","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":false},{"id":"7f7d2b94-0464-4655-9101-9e35ba08a6d9","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":false},{"id":"94383ec2-5a77-4497-82c6-5222b53e1afa","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"17bb4c0f-be3d-4d9c-b63b-32027a4e02e7","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"462ffa14-dd63-437a-bbe4-401e81915588","difficultyBand":"easy","calculatorUse":null,"totalMarks":2,"questionType":"LA","isCalculation":true},{"id":"47d4fc03-078d-4f70-8ef3-9176a2d6ccbb","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"4def930c-1b04-4af7-99ac-fd821d5ba6d0","difficultyBand":"easy","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"713dd78d-e623-4d3b-9df0-3315202a859a","difficultyBand":"medium","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"7dadabbf-e4ad-4909-bf33-fd45bffdbfca","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"8081f0f9-1174-4cd4-abe3-673dc009566a","difficultyBand":"easy","calculatorUse":null,"totalMarks":2,"questionType":"LA","isCalculation":true},{"id":"9768deb2-67df-4442-88c8-1ff2002600a0","difficultyBand":"easy","calculatorUse":null,"totalMarks":2,"questionType":"LA","isCalculation":false},{"id":"a61aac95-7c41-4fc0-86b3-d5ccb35269e7","difficultyBand":"easy","calculatorUse":null,"totalMarks":2,"questionType":"LA","isCalculation":true},{"id":"a7c40a1c-951d-4dc4-b82d-73cd6f655aa1","difficultyBand":"medium","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"b614faa8-39fa-45b9-b23e-e2951b0dacf9","difficultyBand":"easy","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"c44a8baf-e0b0-43c3-9ece-a87049a02101","difficultyBand":"easy","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"d622bdbc-28f3-495b-a6c1-2e8d292ba9b7","difficultyBand":"easy","calculatorUse":null,"totalMarks":2,"questionType":"LA","isCalculation":true},{"id":"da30433f-5ca6-4678-a565-5ebcfdbad0af","difficultyBand":"easy","calculatorUse":null,"totalMarks":2,"questionType":"LA","isCalculation":true}],"title":"1.9.10 Differentiate e^kx, a^kx and ln x (A-level only)","topicId":"9fd87c8e-6997-4807-a05a-03a484b743ce","topicSidebar":"$L3d","topicSlug":"differentiate-e-kx-a-kx-and-ln-x","topicTitle":"1.9.10 Differentiate e^kx, a^kx and ln x (A-level only)"}]}],"$L3e","$L3f"]}]

1.9.10 Differentiate e^kx, a^kx and ln x (A-level only)

Question 3

1.9.10 Differentiate e^kx, a^kx and ln x (A-level only) Questions

1.9.10 Differentiate e^kx, a^kx and ln x (A-level only)

Question 3

1.9.10 Differentiate e^kx, a^kx and ln x (A-level only) Questions