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

Answers (1)

Answer:

-\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

 

 

 

Posted by

infoexpert26

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