Implicit Differentiation for Curve Classification

00:00 8 marks

Consider the curve defined implicitly by the equation \(x^2 y + \sin(y) = x^3\).

(a) Differentiate implicitly to find \(f'(x)\) in terms of \(x\) and \(y\).

(b) Using your expression for \(f'(x)\), find the second derivative \(f”(x)\) and explain how it can be used to classify a turning point on the curve.

Part a

4 minutes 3 marks

Question

Differentiate the equation \(x^2 y + \sin(y) = x^3\) implicitly to find an expression for \(f'(x)\) in terms of \(x\) and \(y\).

Hint Hint

Hint

Remember to apply the product rule to \(x^2 y\) and the chain rule to \(\sin(y)\).

Answer Answer
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Explanation Explanation
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Answer Answer

Complete Answer

\(f'(x) = \frac{3x^2 – 2xy}{x^2 + \cos(y)}\)

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Part b

6 minutes 5 marks

Question

Differentiate your expression for \(f'(x)\) to find the second derivative \(f”(x)\) in terms of \(x\), \(y\), and \(y’\), and explain how it can be used to classify a turning point on the curve.

Hint Hint

Hint

Use the quotient rule on \(f'(x) = \frac{3x^2 – 2xy}{x^2 + \cos(y)}\) and substitute your expression for \(y’\) where necessary.

Answer Answer
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Explanation Explanation
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Answer Answer

Complete Answer

\(f”(x) = \frac{(x^2+\cos(y))(6x-2y-2xy’) – (3x^2-2xy)(2x-\sin(y)y’)}{(x^2+\cos(y))^2}\)

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