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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Aicharya · Accepted Answer

Given that, $y=\cos (\sin x)$ $\begin{aligned} & y_1=-\sin (\sin x) \cdot \cos x \ & y_2=-\cos (\sin x) \cdot \cos ^2 x+\sin (\sin x) \sin x\end{aligned}$ Now, $y_1 \sin x+y_2 \cos x$ $\begin{aligned} & =-\sin (\sin x) \sin x \cos x \ & \quad-y \cos ^3 x+\sin (\sin x) \sin x \cos x \ & =-y \cos ^3 x\end{aligned}$

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

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