#### Provide solution for RD Sharma maths class 12 chapter Differentiation exercise 10.8 question 18

Hint: $\text { Let } x=\cos \theta$

Given: $\sin ^{-1} \sqrt{1-x^{2}} \text { w.r.t } \cot ^{-1}\left(\frac{x}{\sqrt{1-x^{2}}}\right)$

Explanation:

$\text { Let } x=\cos \theta$

\begin{aligned} &u=\sin ^{-1} \sqrt{1-x^{2}} \\\\ &=\sin ^{-1} \sqrt{1-\cos ^{2} \theta} \\\\ &=\sin ^{-1}(\sin \theta) \end{aligned}

\begin{aligned} &v=\cot ^{-1}\left(\frac{x}{\sqrt{1-x^{2}}}\right) \\\\ &=\cot ^{-1}\left(\frac{\cos \theta}{\sqrt{1-\cos ^{2} \theta}}\right) \end{aligned}

$=\cot ^{-1}\left(\frac{\cos \theta}{\sin \theta}\right)=\cot ^{-1}(\cot \theta)$

Now

\begin{aligned} &0

$\begin{array}{ll} u=\sin ^{-1}(\sin \theta)=\theta & \quad \theta \in\left(0, \frac{\pi}{2}\right) \\\\ v=\cot ^{-1}(\cot \theta)=\theta & 2 \theta \in\left(0, \frac{\pi}{2}\right) \end{array}$

\begin{aligned} &u=\cos ^{-1} x \quad\left[\begin{array}{l} x=\cos \theta \\ \theta=\cos ^{-1} x \end{array}\right] \\ &\frac{d u}{d x}=\frac{-1}{\sqrt{1-x^{2}}} \end{aligned}

\begin{aligned} &v=\cos ^{-1} x \\\\ &\frac{d v}{d x}=\frac{-1}{\sqrt{1-x^{2}}} \end{aligned}

$\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{\frac{-1}{\sqrt{1-x^{2}}}}{\frac{-1}{\sqrt{1-x^{2}}}}=1$

$\frac{d u}{d v}=1$

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