#### Provide Solution For R.D.Sharma Maths Class 12 Chapter 10 Differentiation Exercise 10.3 Question 29 Maths Textbook Solution.

Answer:$\frac{dy}{dx}=\frac{a}{a^{2}+x^{2}}$

Hint:

\begin{aligned} &\frac{d}{d x}(\text { constan } t)=0 ; \\ &\frac{d}{d x}\left(x^{n}\right)=n x^{n-1} \end{aligned}

Given:

$\tan ^{-1}\left ( \frac{x-a}{x+a} \right )$

Solution:

Let,

\begin{aligned} &y=\tan ^{-1}\left(\frac{x-a}{x+a}\right) \\ &y=\tan ^{-1}\left(\frac{\frac{x-a}{x}}{\frac{x+a}{x}}\right) \end{aligned}

\begin{aligned} &y=\tan ^{-1}\left(\frac{\frac{x}{x}-\frac{a}{x}}{\frac{x}{x}+\frac{a}{x}}\right) \\ &y=\tan ^{-1}\left(\frac{1-\frac{a}{x}}{1+1 \times \frac{a}{x}}\right) \\ &y=\tan ^{-1}(1)+\tan ^{-1}\left(\frac{a}{x}\right) \end{aligned}

Using,

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

Differentiating its with respect to $x$ using chain tule.

\begin{aligned} &\frac{d y}{d x}=0-\frac{1}{1+\left(\frac{a}{x}\right)^{2}} \frac{d}{d x}\left(\frac{a}{x}\right) \\ &\frac{d y}{d x}=\frac{-1}{1+\frac{a^{2}}{x^{2}}}\left(-\frac{a}{x^{2}}\right) \\ &\frac{d y}{d x}=\frac{-1}{\frac{x^{2}+a^{2}}{x^{2}}} \times\left(\frac{-a}{x^{2}}\right) \\ &\frac{d y}{d x}=\frac{-x^{2}}{x^{2}+a^{2}} \times\left(\frac{-a}{x^{2}}\right) \end{aligned}                                                                                                            Using   $\frac{d}{dx}\left ( x^{n} \right )=nx^{n-1}$

$\frac{d}{dx}=\frac{a}{x^{2}+a^{2}}$                                                                                                                    $\frac{d}{dx}\left ( constant \right )=0$