#### Please solve RD Sharma class 12 chapter Differentiation exercise 10.4 question 13 maths textbook solution

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

Hint:

Use trigonometric identities and differentiation formula of inverse trigonometric functions

Given:

$y \sqrt{1-x^{2}}+x \sqrt{1-y^{2}}=1$

Solution:

Let $x=\sin A, y=\sin B$

$\therefore$ The given equation becomes

$\sin B \sqrt{1-\sin ^{2} A}+\sin A \sqrt{1-\sin ^{2} B}=1$

$\sin B \sqrt{\cos ^{2} A}+\sin A \sqrt{\cos ^{2} B}=1 \quad\left[\because \sin ^{2} \theta+\cos ^{2} \theta=1\right]$

$\sin B \cos A+\sin A \cos B=1$

$\sin (A+B)=1 \quad[\because \sin (A+B)=\sin A \cos B+\cos A \sin B]$

\begin{aligned} &A+B=\sin ^{-1}(1) \\ &A+B=\frac{\pi}{2} \end{aligned}

$\therefore \sin ^{-1} x+\sin ^{-1} y=\frac{\pi}{2} \quad[\because x=\sin A, y=\sin B]$

Differentiate $\left(\sin ^{-1} x+\sin ^{-1} y=\frac{\pi}{2}\right)$ w.r.t x

$\frac{d\left(\sin ^{-1} x+\sin ^{-1} y\right)}{d x}=\frac{d\left(\frac{\pi}{2}\right)}{d x}$

$\frac{d\left(\sin ^{-1} x\right)}{d x}+\frac{d\left(\sin ^{-1} y\right)}{d x}=0 \quad\left[\because \frac{d(\operatorname{cons} \tan t)}{d x}=0\right]$

$\frac{1}{\sqrt{1-x^{2}}}+\frac{d\left(\sin ^{-1} y\right)}{d y} \times \frac{d y}{d x}=0 \quad\left[\because \frac{d\left(\sin ^{-1} x\right)}{d x}=\frac{1}{\sqrt{1-x^{2}}}\right]$

$\frac{1}{\sqrt{1-x^{2}}}+\frac{1}{\sqrt{1-y^{2}}} \cdot \frac{d y}{d x}=0$

$\frac{1}{\sqrt{1-y^{2}}} \cdot \frac{d y}{d x}=-\frac{1}{\sqrt{1-x^{2}}}$

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

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

Hence if   $y \sqrt{1-x^{2}}+x \sqrt{1-y^{2}}=1$

Then, $\frac{d y}{d x}=-\sqrt{\frac{1-y^{2}}{1-x^{2}}}$  Is the required answer

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