Let f(x)=e^x, g(x)=sin ^{-1} x and h(x)=f(g(x | vedclass

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(C) Given $f(x)=e^x$ and $g(x)=\sin ^{-1} x$. $h(x)=f(g(x))=e^{\sin ^{-1} x}$. Differentiating $h(x)$ with respect to $x$: $h^{\prime}(x)=\frac{d}{dx}

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$\frac{1}{1-x^2}$

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(C) Given $f(x)=e^x$ and $g(x)=\sin ^{-1} x$. $h(x)=f(g(x))=e^{\sin ^{-1} x}$. Differentiating $h(x)$ with respect to $x$: $h^{\prime}(x)=\frac{d}{dx}(e^{\sin ^{-1} x}) = e^{\sin ^{-1} x} \cdot \frac{d}{dx}(\sin ^{-1} x) = e^{\sin ^{-1} x} \cdot \frac{1}{\sqrt{1-x^2}}$. Now,consider the ratio $\frac{h^{\prime}(x)}{h(x)}$: $\frac{h^{\prime}(x)}{h(x)} = \frac{e^{\sin ^{-1} x} \cdot \frac{1}{\sqrt{1-x^2}}}{e^{\sin ^{-1} x}} = \frac{1}{\sqrt{1-x^2}}$. Finally,squaring the result: $\left(\frac{h^{\prime}(x)}{h(x)}\right)^2 = \left(\frac{1}{\sqrt{1-x^2}}\right)^2 = \frac{1}{1-x^2}$.

Let $f(x)=e^x, g(x)=\sin ^{-1} x$ and $h(x)=f(g(x))$,then $\left(\frac{h^{\prime}(x)}{h(x)}\right)^2$ is equal to

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