we prepare for explicit differentiation.*"},{"type":"text","order":1,"title":"Differentiate $y$ explicitly","content":"Now we differentiate $y = \\cos^{-1}\\left(\\frac{1}{2} - \\sin x\\right)$ with respect to $x$ using the chain rule.\n\nRecall the standard derivative for inverse cosine:\n\n$$\\frac{d}{du}\\left(\\cos^{-1} u\\right) = -\\frac{1}{\\sqrt{1-u^2}}$$\n\nLetting $u = \\frac{1}{2} - \\sin x$, its derivative with respect to $x$ is:\n\n$$\\frac{du}{dx} = -\\cos x$$\n\nApplying the chain rule, $\\frac{dy}{dx} = \\frac{dy}{du} \\times \\frac{du}{dx}$:\n\n$$\\begin{aligned} \\frac{dy}{dx} &= -\\frac{1}{\\sqrt{1 - \\left(\\frac{1}{2} - \\sin x\\right)^2}} \\cdot (-\\cos x) \\\\ \\frac{dy}{dx} &= \\frac{\\cos x}{\\sqrt{1 - \\left(\\frac{1}{2} - \\sin x\\right)^2}} \\end{aligned}$$"},{"type":"text","order":2,"title":"Set the derivative to zero","content":"To find the coordinates where the gradient is zero, we set $\\frac{dy}{dx} = 0$:\n\n$$\\frac{\\cos x}{\\sqrt{1 - \\left(\\frac{1}{2} - \\sin x\\right)^2}} = 0$$\n\nThis fraction is zero when its numerator is zero, provided the denominator is non-zero:\n\n$$\\cos x = 0$$"},{"type":"text","order":3,"title":"Solve for $x$","content":"Within the given interval $0 \\le x \\le \\pi$, we solve:\n\n$$\\cos x = 0 \\implies x = \\frac{\\pi}{2}$$\n\n*(This earns the first method mark **M1**)*"},{"type":"text","order":4,"title":"Find the $y$-coordinate","content":"Now substitute $x = \\frac{\\pi}{2}$ back into our explicit equation for $y$ to find the corresponding $y$-coordinate:\n\n$$\\begin{aligned} y &= \\cos^{-1}\\left(\\frac{1}{2} - \\sin\\left(\\frac{\\pi}{2}\\right)\\right) \\\\ y &= \\cos^{-1}\\left(\\frac{1}{2} - 1\\right) \\\\ y &= \\cos^{-1}\\left(-\\frac{1}{2}\\right) \\end{aligned}$$\n\n*(This substitution earns **M1** and solving for $\\cos y = -\\frac{1}{2}$ earns **M1**)*"},{"type":"text","order":5,"title":"State final coordinates","content":"Since $0 \\le y \\le \\pi$, the principal value for our inverse cosine is:\n\n$$y = \\frac{2\\pi}{3}$$\n\nTherefore, the coordinates of the point where $\\frac{dy}{dx} = 0$ are:\n\n$$\\left(\\frac{\\pi}{2}, \\frac{2\\pi}{3}\\right)$$\n\n*(This final solution earns the accuracy mark **A1**)*"}]}],"generatedAt":"2026-06-09T05:22:24.381Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null},{"id":"4eedb155-97aa-466c-9eba-76d75f94f3f5","revisionId":"034b1e96-d9e5-450b-8395-819d6521d6d5","questionNumber":null,"questionType":"LA","specification":"A curve has the equation $2x^2 + xy - y^2 = 5$:","difficulty":1,"difficultyBand":"medium","totalMarks":8,"sourceTitle":null,"calculatorUse":null,"parts":[{"id":"4b8891c4-483f-4fa5-ae89-7297df8cc574","label":"a","content":"Find an expression for $\\frac{dy}{dx}$","markscheme":"- Differentiates $2x^2$ to $4x$ and $y^2$ to $2y \\frac{dy}{dx}$ M1\n- Applies product rule to $xy$ to get $y + x \\frac{dy}{dx}$ M1\n- Sets the derivative of the constant 5 to 0 B1\n- Rearranges to factorise $\\frac{dy}{dx}$ M1\n- Correct final expression $\\frac{dy}{dx} = \\frac{4x + y}{2y - x}$ A1","marks":5,"order":0,"markingRubric":{"totalMarks":5,"points":[{"id":"p1","marks":1,"description":"Implicit differentiation of squared terms.","code":"M"},{"id":"p2","marks":1,"description":"Implicit differentiation of the product term $xy$.","code":"M"},{"id":"p3","marks":1,"description":"Constant term differentiates to zero.","code":"B"},{"id":"p4","marks":1,"description":"Correct algebra to isolate $dy/dx$.","code":"M"},{"id":"p5","marks":1,"description":"Correct final expression.","code":"A"}],"finalAnswers":["\\frac{dy}{dx} = \\frac{4x + y}{2y - x}"],"generatedAt":"2026-05-06T14:04:19.108Z","modelVersion":"gemini-3-flash-preview"},"explanation":{"methods":[{"id":"implicit_differentiation","label":"Direct Implicit Differentiation","steps":[{"type":"text","order":0,"title":"Identify the goal","content":"We need to find an expression for $\\frac{dy}{dx}$ for the given curve:\n\n$$2x^2 + xy - y^2 = 5$$\n\nTo

Differentiation

Question 44

Differentiation Questions

Differentiation

Question 44

Differentiation Questions