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

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

Answers (1)

best_answer

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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