If y_k is the k-th derivative of y with respe | vedclass

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(D) Given that $y = \cos(\sin x)$.First derivative $y_1 = \frac{dy}{dx} = -\sin(\sin x) \cdot \cos x$.Second derivative $y_2 = \frac{d^2y}{dx^2} = \fr

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$-y \cos^3 x$

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(D) Given that $y = \cos(\sin x)$.First derivative $y_1 = \frac{dy}{dx} = -\sin(\sin x) \cdot \cos x$.Second derivative $y_2 = \frac{d^2y}{dx^2} = \frac{d}{dx}[-\sin(\sin x) \cdot \cos x]$.Using the product rule: $y_2 = -[\cos(\sin x) \cdot \cos x \cdot \cos x + \sin(\sin x) \cdot (-\sin x)]$.$y_2 = -\cos(\sin x) \cos^2 x + \sin(\sin x) \sin x$.Now, calculate $y_1 \sin x + y_2 \cos x$:$y_1 \sin x + y_2 \cos x = [-\sin(\sin x) \cos x] \sin x + [-\cos(\sin x) \cos^2 x + \sin(\sin x) \sin x] \cos x$.$= -\sin(\sin x) \sin x \cos x - \cos(\sin x) \cos^3 x + \sin(\sin x) \sin x \cos x$.$= -\cos(\sin x) \cos^3 x$.Since $y = \cos(\sin x)$, the expression becomes $-y \cos^3 x$.

If $y_k$ is the $k$-th derivative of $y$ with respect to $x$, and $y = \cos(\sin x)$, then $y_1 \sin x + y_2 \cos x$ is equal to

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