Need Solution for R.D.Sharma Maths Class 12 Chapter 10 Differentiation Exercise 10.3 Question 26 Maths Textbook Solution.

Answer:$\frac{1}{2\sqrt{x}\left ( 1+x \right )}$

Hint:

$\frac{d}{dx}\left ( constant \right )=0;$

$\frac{d}{d x}\left ( x^{n} \right )=nx^{n-1}$

Given:

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

Solution:

Let,

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

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

$y=\tan ^{-1}\sqrt{x}+\tan ^{-1}\sqrt{a}$

Differentiating it with respect to x using chain qule.

$\frac{dy}{dx}=\frac{d}{dx}\left ( \tan ^{-1}\sqrt{x} \right )+\frac{d}{dx}\left ( \tan ^{-1}\sqrt{a} \right )$

Since ,

\begin{aligned} &\frac{\mathrm{d}}{\mathrm{dx}}\left(\tan ^{-1} \mathrm{x}\right)=\frac{1}{1+\mathrm{x}^{2}} ; \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \mathrm{x}^{\mathrm{n}-1} \\ &\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{1+(\sqrt{\mathrm{x}})^{2}} \frac{\mathrm{d}}{\mathrm{dx}}(\sqrt{\mathrm{x}})+0 \\ &\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{1+\left(\mathrm{x}^{1 / 2}\right)^{2}} \times \frac{1}{2}(\mathrm{x})^{\frac{1}{2}-1} \\ &\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{1+\mathrm{x}} \times \frac{1}{2}(\mathrm{x})^{-\frac{1}{2}} \\ &\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{1+\mathrm{x}} \times \frac{1}{2 \sqrt{\mathrm{x}}} \\ &\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2 \sqrt{\mathrm{x}}(1+\mathrm{x})} \end{aligned}