If y = sin [cos (sin x)], then frac{dy}{dx} = | vedclass

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(A) Given $y = \sin [\cos (\sin x)]$.Applying the chain rule, we differentiate with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx} (\sin [\cos (\sin x)]

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given $y = \sin [\cos (\sin x)]$.Applying the chain rule, we differentiate with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx} (\sin [\cos (\sin x)]) = \cos [\cos (\sin x)] \cdot \frac{d}{dx} (\cos (\sin x))$.Now, differentiate $\cos (\sin x)$:$\frac{d}{dx} (\cos (\sin x)) = -\sin (\sin x) \cdot \frac{d}{dx} (\sin x)$.Finally, differentiate $\sin x$:$\frac{d}{dx} (\sin x) = \cos x$.Combining all these parts:$\frac{dy}{dx} = \cos [\cos (\sin x)] \cdot [-\sin (\sin x)] \cdot \cos x = -\cos [\cos (\sin x)] \sin (\sin x) \cos x$.

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

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