If y = e^{cos ^{-1}left(sqrt{1-x^2}right)},th | vedclass

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(C) Given $y = e^{\cos ^{-1}\left(\sqrt{1-x^2}\right)}$.Taking the natural logarithm on both sides,we get $\log _e y = \cos ^{-1} \sqrt{1-x^2}$.Differ

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Given $y = e^{\cos ^{-1}\left(\sqrt{1-x^2}\right)}$.Taking the natural logarithm on both sides,we get $\log _e y = \cos ^{-1} \sqrt{1-x^2}$.Differentiating both sides with respect to $x$:$\frac{1}{y} \frac{d y}{d x} = \frac{d}{d x} \left( \cos ^{-1} \sqrt{1-x^2} \right)$.Using the chain rule,$\frac{d}{d x} \cos ^{-1}(u) = \frac{-1}{\sqrt{1-u^2}} \frac{du}{dx}$,where $u = \sqrt{1-x^2}$.$\frac{1}{y} \frac{d y}{d x} = \frac{-1}{\sqrt{1-(1-x^2)}} \cdot \frac{d}{dx} \left( (1-x^2)^{1/2} \right)$.$\frac{1}{y} \frac{d y}{d x} = \frac{-1}{\sqrt{x^2}} \cdot \left( \frac{1}{2} (1-x^2)^{-1/2} \cdot (-2x) \right)$.$\frac{1}{y} \frac{d y}{d x} = \frac{-1}{|x|} \cdot \frac{-x}{\sqrt{1-x^2}}$.Assuming $x > 0$,$|x| = x$,so $\frac{1}{y} \frac{d y}{d x} = \frac{x}{x \sqrt{1-x^2}} = \frac{1}{\sqrt{1-x^2}}$.

If $y = e^{\cos ^{-1}\left(\sqrt{1-x^2}\right)}$,then find $\frac{1}{y} \frac{d y}{d x}$.

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