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If $f(x)=\sin (\sin x)$ and $f^{\prime \prime}(x)+\tan x f^{\prime}(x)+g(x)$ $=0$, then $g(x)$ is :
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Verified Answer
The correct answer is:
$\cos ^2 x \sin (\sin x)$
$\cos ^2 x \sin (\sin x)$
$$
\begin{aligned}
& \text { } f(x)=\sin (\sin x) \\
& \Rightarrow f^{\prime}(x)=\cos (\sin x) \cdot \cos x \\
& \Rightarrow f^{\prime \prime}(x)=-\sin (\sin x) \cdot \cos ^2 x+\cos (\sin x) \text {. } \\
& (-\sin x) \\
& =-\cos ^2 x \cdot \sin (\sin x)-\sin x \cdot \cos (\sin x) \\
& \text { Now } f^{\prime \prime}(x)+\tan x . f^{\prime}(x)+g(x)=0 \\
& \Rightarrow g(x)=\cos ^2 x \cdot \sin (\sin x)+\sin x \cdot \cos (\sin x) \\
& -\tan x \cdot \cos x \cdot \cos (\sin x) \\
& \Rightarrow g(x)=\cos ^2 x \cdot \sin (\sin x) \text {. } \\
&
\end{aligned}
$$
\begin{aligned}
& \text { } f(x)=\sin (\sin x) \\
& \Rightarrow f^{\prime}(x)=\cos (\sin x) \cdot \cos x \\
& \Rightarrow f^{\prime \prime}(x)=-\sin (\sin x) \cdot \cos ^2 x+\cos (\sin x) \text {. } \\
& (-\sin x) \\
& =-\cos ^2 x \cdot \sin (\sin x)-\sin x \cdot \cos (\sin x) \\
& \text { Now } f^{\prime \prime}(x)+\tan x . f^{\prime}(x)+g(x)=0 \\
& \Rightarrow g(x)=\cos ^2 x \cdot \sin (\sin x)+\sin x \cdot \cos (\sin x) \\
& -\tan x \cdot \cos x \cdot \cos (\sin x) \\
& \Rightarrow g(x)=\cos ^2 x \cdot \sin (\sin x) \text {. } \\
&
\end{aligned}
$$
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