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

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(B) Given $f(x) = e^x$ and $g(x) = \sin^{-1}x$. Since $h(x) = f(g(x))$,we have $h(x) = f(\sin^{-1}x) = e^{\sin^{-1}x}$. Now,differentiate $h(x)$ with

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

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(B) Given $f(x) = e^x$ and $g(x) = \sin^{-1}x$. Since $h(x) = f(g(x))$,we have $h(x) = f(\sin^{-1}x) = e^{\sin^{-1}x}$. Now,differentiate $h(x)$ with respect to $x$ using the chain rule: $h'(x) = \frac{d}{dx}(e^{\sin^{-1}x}) = e^{\sin^{-1}x} \cdot \frac{d}{dx}(\sin^{-1}x) = e^{\sin^{-1}x} \cdot \frac{1}{\sqrt{1 - x^2}}$. Therefore,the ratio $h'(x)/h(x)$ is: $\frac{h'(x)}{h(x)} = \frac{e^{\sin^{-1}x} \cdot \frac{1}{\sqrt{1 - x^2}}}{e^{\sin^{-1}x}} = \frac{1}{\sqrt{1 - x^2}}$.

Let $f(x) = e^x$,$g(x) = \sin^{-1}x$ and $h(x) = f(g(x))$,then $h'(x)/h(x) = $

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