If y = sin(sin x) and y'' + f(x) cdot y' + g( | vedclass

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(A) Given $y = \sin(\sin x)$ . . . $(i)$Differentiating with respect to $x$ using the chain rule:$y' = \cos(\sin x) \cdot \cos x$ . . . $(ii)$Differen

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$\frac{1}{2} \sin(2x)$

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(A) Given $y = \sin(\sin x)$ . . . $(i)$Differentiating with respect to $x$ using the chain rule:$y' = \cos(\sin x) \cdot \cos x$ . . . $(ii)$Differentiating again with respect to $x$:$y'' = -\sin(\sin x) \cdot \cos^2 x - \sin x \cdot \cos(\sin x)$From equation $(ii)$,we have $\cos(\sin x) = \frac{y'}{\cos x}$. Substituting this into the expression for $y''$:$y'' = -\sin(\sin x) \cdot \cos^2 x - \sin x \cdot \left(\frac{y'}{\cos x}ight)$$y'' = -y \cdot \cos^2 x - 	an x \cdot y'$Rearranging the terms gives:$y'' + 	an x \cdot y' + \cos^2 x \cdot y = 0$Comparing this with $y'' + f(x) \cdot y' + g(x) \cdot y = 0$,we get $f(x) = 	an x$ and $g(x) = \cos^2 x$.Therefore,$f(x) \cdot g(x) = 	an x \cdot \cos^2 x = \frac{\sin x}{\cos x} \cdot \cos^2 x = \sin x \cdot \cos x$.Using the identity $\sin(2x) = 2 \sin x \cos x$,we get:$f(x) \cdot g(x) = \frac{1}{2} \sin(2x)$.

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

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