#### Please solve RD Sharma class 12 chapter Differentiation exercise 10.2 question 32 maths textbook solution

Answer:  $\frac{-2\left(x^{2}-1\right)}{x^{4}+x^{2}+1}$

Hint: You must know about the rules of solving derivative of logarithm functions.

Given: $\log \left(\frac{x^{2}+x+1}{x^{2}-x+1}\right)$

Solution:

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

Differentiate with respect to x,

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

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

Now apply quotient rule,  $\frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}$

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

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

$\frac{d y}{d x}=\left(\frac{x^{2}-x+1}{x^{2}+x+1}\right)\left[\frac{\left(2 x^{3}-2 x^{2}+2 x+x^{2}-x+1-2 x^{3}-2 x^{2}-2 x+x^{2}+x+1\right)}{\left(x^{2}-x+1\right)^{2}}\right]$

$\frac{d y}{d x}=\frac{-4 x^{2}+2 x^{2}+2}{\left(x^{2}+x+1\right)\left(x^{2}-x+1\right)}$

$\frac{d y}{d x}=\frac{-4 x^{2}+2 x^{2}+2}{\left(x^{2}+1\right)^{2}-(x)^{2}}$

\begin{aligned} &\frac{d y}{d x}=\frac{-2\left(x^{2}-1\right)}{x^{4}+1+2 x^{2}-x^{2}} \\ &\frac{d y}{d x}=\frac{-2\left(x^{2}-1\right)}{x^{4}+x^{2}+1} \end{aligned}