#### Need Solution for R.D.Sharma Maths Class 12 Chapter 10 Differentiation Exercise 10.3 Question 33 Maths Textbook Solution.

Answer: $\frac{dy}{dx}=\frac{1}{3x^{2}\left ( 1+x^{2}3 \right )}$

Hint:

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

Given:

$\tan ^{-1}\left[\frac{\mathrm{x}^{1 / 3}+\mathrm{a}^{1 / 3}}{1-(\mathrm{ax})^{1 / 3}}\right]$

Solution:

Let,

\begin{aligned} &y=\tan ^{-1}\left[\frac{x^{13}+a^{1 / 3}}{1-(a x)^{1 / 3}}\right] \\ &y=\tan ^{-1}\left(x^{1 / 3}\right)+\tan ^{-1}\left(a^{1 / 3}\right) \\ &\text { Since, } \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right) \end{aligned}

Differentiating it with respect to x using chain rule,

\begin{aligned} &\frac{d y}{d x}=\frac{1}{1+\left(x^{1 / 3}\right)^{2}} \frac{d}{d x}\left(x^{1 / 3}\right)+\frac{1}{1+\left(a^{1 / 3}\right)^{2}} \frac{d}{d x}\left(a^{1 / 3}\right) \\ &\frac{d y}{d x}=\frac{1}{1+x^{2 / 3}} \times\left(\frac{1}{3} x^{\frac{1}{3}-1}\right)+\frac{1}{1+\left(a^{1 / 3}\right)^{2}} \times 0 \\ &\frac{d y}{d x}=\frac{1}{1+x^{2 / 3}} \times \frac{1}{3} x^{-2 / 3} \\ &\frac{d y}{d x}=\frac{1}{3 x^{2 / 3}\left(1+x^{2 / 3}\right)} \end{aligned}

Using

\begin{aligned} &\frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^{2}} \\ &\frac{d}{d x}\left(x^{n}\right)=n x^{n-1} \\ &\frac{d}{d x}(\text { constant })=0 \end{aligned}