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

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(A) Given $f(x) = \cos^{-1} x$ and $g(x) = e^x$. We define $h(x) = g(f(x)) = e^{\cos^{-1} x}$. Now,we differentiate $h(x)$ with respect to $x$ using t

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

Vedclass · Answer

(A) Given $f(x) = \cos^{-1} x$ and $g(x) = e^x$. We define $h(x) = g(f(x)) = e^{\cos^{-1} x}$. Now,we differentiate $h(x)$ with respect to $x$ using the chain rule: $h'(x) = \frac{d}{dx}(e^{\cos^{-1} x}) = e^{\cos^{-1} x} \cdot \frac{d}{dx}(\cos^{-1} x)$. Since $\frac{d}{dx}(\cos^{-1} x) = \frac{-1}{\sqrt{1-x^2}}$,we have: $h'(x) = e^{\cos^{-1} x} \cdot \left( \frac{-1}{\sqrt{1-x^2}} \right)$. Now,we calculate the ratio $\frac{h'(x)}{h(x)}$: $\frac{h'(x)}{h(x)} = \frac{e^{\cos^{-1} x} \cdot \left( \frac{-1}{\sqrt{1-x^2}} \right)}{e^{\cos^{-1} x}} = \frac{-1}{\sqrt{1-x^2}}$. Thus,the correct option is $A$.

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

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