If f(x) = sin(sin x) and f''(x) + tan x f'(x) | vedclass

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(D) Given $f(x) = \sin(\sin x)$.First,find the first derivative $f'(x)$ using the chain rule:$f'(x) = \cos(\sin x) \cdot \cos x$.Next,find the second

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$\cos^2 x \sin(\sin x)$

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(D) Given $f(x) = \sin(\sin x)$.First,find the first derivative $f'(x)$ using the chain rule:$f'(x) = \cos(\sin x) \cdot \cos x$.Next,find the second derivative $f''(x)$ using the product rule and chain rule:$f''(x) = \frac{d}{dx}[\cos(\sin x) \cdot \cos x]$$f''(x) = [-\sin(\sin x) \cdot \cos x] \cdot \cos x + \cos(\sin x) \cdot (-\sin x)$$f''(x) = -\cos^2 x \sin(\sin x) - \sin x \cos(\sin x)$.Now,substitute $f'(x)$ and $f''(x)$ into the given equation $f''(x) + 	an x f'(x) + g(x) = 0$:$-\cos^2 x \sin(\sin x) - \sin x \cos(\sin x) + 	an x [\cos(\sin x) \cos x] + g(x) = 0$.Since $	an x \cos x = \sin x$,the equation becomes:$-\cos^2 x \sin(\sin x) - \sin x \cos(\sin x) + \sin x \cos(\sin x) + g(x) = 0$.The terms $-\sin x \cos(\sin x)$ and $+\sin x \cos(\sin x)$ cancel out:$-\cos^2 x \sin(\sin x) + g(x) = 0$.Therefore,$g(x) = \cos^2 x \sin(\sin x)$.

If $f(x) = \sin(\sin x)$ and $f''(x) + \tan x f'(x) + g(x) = 0$,then $g(x)$ is

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