Question

If $y=\sin \left[\cos \left(\sin x\right)\right]$, then $\frac{dy}{dx}=$

• A. $-\cos \left[\cos \left(\sin x\right)\right]\sin \left(\cos x\right)\cos x$
• B. $-\cos \left[\cos \left(\sin x\right)\right]\sin \left(\sin x\right)\cos x$
• C. $\cos \left[\cos \left(\sin x\right)\right]\sin \left(\cos x\right)\cos x$
• D. $\cos \left[\cos \left(\sin x\right)\right]\sin \left(\sin x\right)\cos x$

Answer 1

The correct option is B

$-\cos \left[\cos \left(\sin x\right)\right]\sin \left(\sin x\right)\cos x$

Explanation for the correct option:

Finding the value of $\frac{dy}{dx}$:

Given that,

$y=\sin \left[\cos \left(\sin x\right)\right]$

Using Chain Rule, Differentiating with respect to $x$ both sides:

$$\begin{gathered} \frac{dy}{dx}=\cos \left[\cos (\sin x)\right]\times \frac{d}{dx}\left[\cos (\sin x)\right]\left[\because \frac{d\left(u·v\right)}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}\right] \\ \Rightarrow \frac{dy}{dx}=\cos \left[\cos (\sin x)\right]\times \left[-\sin (\sin x)\right]\times \frac{d}{dx}(\sin x)\left[\because \frac{d\left(\sin x\right)}{dx}=\cos x,\frac{d\left(\cos x\right)}{dx}=-\sin x\right] \\ \Rightarrow \frac{dy}{dx}=-\cos \left[\cos (\sin x)\right]\sin (\sin x)\cos x \end{gathered}$$

Hence, Option (B) is the correct answer.