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  • need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.4 question 19

need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.4 question 19

Answers (1)

Answer:

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

Hint:

Use differentiation formulas

Given:

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

Solution:

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

\tan a=\frac{x^{2}-y^{2}}{x^{2}+y^{2}}

\begin{aligned} &\left(x^{2}+y^{2}\right) \tan a=x^{2}-y^{2} \\ &x^{2}-y^{2}=(\tan a)\left(x^{2}+y^{2}\right) \end{aligned}

Differentiate the given equation w.r.t x

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

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

2 x-\frac{d y^{2}}{d y} \times \frac{d y}{d x}=\tan a \cdot\left[2 x+\frac{d y^{2}}{d y} \times \frac{d y}{d x}\right]

2 x-2 y \frac{d y}{d x}=\tan a\left[2 x+2 y \cdot \frac{d y}{d x}\right]

2 x-2 y \frac{d y}{d x}=2 x \tan a+2 y \tan a \frac{d y}{d x}

2 y \tan a \frac{d y}{d x}+2 y \frac{d y}{d x}=2 x-2 x \tan a

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

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

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

Thus, proved

 

Posted by

infoexpert26

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