If $y=\sin (\sin x)$, and $\frac{d^2 y}{d x^2}+\frac{d y}{d x} \tan x+f(x)=0$ then f(x) is :
Answer & explanation
Correct answer: option 4
$\frac{d y}{d x}=\cos (\sin x) . \cos x$
$\Rightarrow \frac{d^2 y}{d x^2}=-\cos (\sin x) . \sin x+\cos x[-\sin (\sin x) \cos x]$
$\Rightarrow \frac{d^2 y}{d x^2}+\frac{d y}{d x} \tan x=-\cos (\sin x) . \sin x-\cos ^2 x . \sin (\sin x)+\cos (\sin x) . \cos x . \tan x=-\cos ^2 x . \sin (\sin x)$
$\Rightarrow f(x)=\cos ^2 x . \sin (\sin x)$
Hence (4) is correct answer.