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

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(B) $h(x) = f(g(x))$$h(x) = f(\sin^{-1} x) = e^{\sin^{-1} x}$Differentiating $h(x)$ with respect to $x$ using the chain rule:$h^{\prime}(x) = \frac{d}

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

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(B) $h(x) = f(g(x))$$h(x) = f(\sin^{-1} x) = e^{\sin^{-1} x}$Differentiating $h(x)$ with respect to $x$ using the chain rule:$h^{\prime}(x) = \frac{d}{dx}(e^{\sin^{-1} x}) = e^{\sin^{-1} x} \cdot \frac{d}{dx}(\sin^{-1} x)$$h^{\prime}(x) = e^{\sin^{-1} x} \cdot \frac{1}{\sqrt{1-x^2}}$Now,calculating 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{h^{\prime}(x)}{h(x)} = \frac{1}{\sqrt{1-x^2}}$

If $f(x)=e^x$,$g(x)=\sin^{-1} x$ and $h(x)=f(g(x))$,then $\frac{h^{\prime}(x)}{h(x)}$ is

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