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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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Verified Answer
The correct answer is:
$\frac{1}{\sqrt{1-x^2}}$
$\begin{aligned} & y=e^{\cos ^{-1}\left(\sqrt{1-x^2}\right)} \\ & \Rightarrow \log _e y=\cos ^{-1} \sqrt{1-x^2}\end{aligned}$
Diff. w.r.t. $\mathrm{x}$
$\frac{1}{y} \cdot \frac{d y}{d x}=\frac{-1}{\sqrt{1-\left(1-x^2\right)}} \times \frac{1}{2 \sqrt{1-x^2}} \times(-2 x)=\frac{2 x}{2 \sqrt{x^2} \sqrt{1-x^2}}=\frac{1}{\sqrt{1-x^2}}$
Diff. w.r.t. $\mathrm{x}$
$\frac{1}{y} \cdot \frac{d y}{d x}=\frac{-1}{\sqrt{1-\left(1-x^2\right)}} \times \frac{1}{2 \sqrt{1-x^2}} \times(-2 x)=\frac{2 x}{2 \sqrt{x^2} \sqrt{1-x^2}}=\frac{1}{\sqrt{1-x^2}}$
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