Q192 / 270 · NDA 2010 (Phase 2) · Mathematics

What is the derivative of $\sin ^2 x$ with respect to $\cos ^2 x$ ?

  • A
    $\tan ^2 x$
  • B
    $\cot ^2 x$
  • Correct answer
    C
    -1
  • D
    1

Solution

$\begin{aligned} & \text { Let } u=\sin ^2 x \\ & \text {and } v=\cos ^2 x \\ & \begin{aligned} \therefore \frac{d u}{d x} & =2 \sin x \cdot \cos x \\ \frac{d u}{d x} & =-2 \cos x \cdot \sin x \\ \therefore \quad \frac{d u}{d v} & =\frac{d u}{d x} \times \frac{d x}{d v} \\ & =\frac{2 \sin x \cdot \cos x}{-2 \sin x \cdot \cos x}=-1\end{aligned}\end{aligned}$

Hence, the correct answer is option: 'C'.

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