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

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(B) 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$

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(B) 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 $dy/dx = $

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