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  • Please solve RD Sharma class 12 chapter Differentiation exercise 10.4 question 9 maths textbook solution

Please solve RD Sharma class 12 chapter Differentiation exercise 10.4 question 9 maths textbook solution

Answers (1)

Answer:\left[-\left(\frac{x}{y}\right)\right]

Hint:

Use chain rule and \frac{d\left(\tan ^{-1} x\right)}{d x}=\frac{1}{1+x^{2}}

Given:

\tan ^{-1}\left(x^{2}+y^{2}\right)=a

Solution:

Differentiate the given equation w.r.t x

\frac{d}{d x}\left(\tan ^{-1}\left(x^{2}+y^{2}\right)\right)=\frac{d(a)}{d x}

\frac{d\left(\tan ^{-1}\left(x^{2}+y^{2}\right)\right)}{d\left(x^{2}+y^{2}\right)} \times \frac{d\left(x^{2}+y^{2}\right)}{d x}=0                [Using chain rule] \left[\because \frac{d(\operatorname{cons} \tan t)}{d x}=0\right]

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

\frac{1}{1+\left(x^{2}+y^{2}\right)^{2}} \times\left(2 x+\frac{d y^{2}}{d y} \times \frac{d y}{d x}\right)=0

\frac{1}{1+\left(x^{2}+y^{2}\right)^{2}} \times\left(2 x+2 y \frac{d y}{d x}\right)=0

\frac{2 x}{1+\left(x^{2}+y^{2}\right)^{2}}+\frac{2 y}{1+\left(x^{2}+y^{2}\right)^{2}} \frac{d y}{d x}=0

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

\frac{d y}{d x}=-\frac{2 x}{\left[1+\left(x^{2}+y^{2}\right)^{2}\right]} \times \frac{1+\left(x^{2}+y^{2}\right)^{2}}{2 y}

\frac{d y}{d x}=\frac{-2 x}{2 y}=-\frac{x}{y}

\frac{d y}{d x}=-\frac{x}{y}

Hence \frac{d y}{d x}=-\frac{x}{y} is the required answer

 

Posted by

infoexpert26

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