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

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Aicharya · Accepted Answer

$\begin{aligned} & y=e^{\cos ^{-1}\left(\sqrt{1-x^2}ight)} \ & \Rightarrow \log _e y=\cos ^{-1} \sqrt{1-x^2}\end{aligned}$ Diff. w.r.t. $\mathrm{x}$ $\frac{1}{y} \cdot \frac{d y}{d x}=\frac{-1}{\sqrt{1-\left(1-x^2ight)}} 	imes \frac{1}{2 \sqrt{1-x^2}} 	imes(-2 x)=\frac{2 x}{2 \sqrt{x^2} \sqrt{1-x^2}}=\frac{1}{\sqrt{1-x^2}}$

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

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