First slide

Derivatives

Question

$If x = 3 \cos θ - 2 \cos^{3} θ, y = 3 \sin θ - 2 \sin^{3} θ then \frac{d y}{d x} =$

Moderate

A

$\tan θ$
B

$\cot θ$
C

$\cot \frac{θ}{2}$
D

$\tan \frac{θ}{2}$

Solution

$\frac{d y}{d x} = \frac{3 \cos θ - 6 \sin^{2} θ \cos θ}{- 3 \sin θ + 6 \cos^{2} θ \sin θ} = \frac{3 \cos θ (1 - 2 \sin^{2} θ)}{3 \sin θ (- 1 + 2 \cos^{2} θ)} = \frac{\cos θ \cdot \cos 2 θ}{\sin θ \cdot \cos 2 θ} = \cot θ$

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