#### Explain solution RD Sharma class 12 chapter 10 Differentiation exercise Multiple choice question 16 maths

$\frac{x^{2}-1}{x^{2}-4}$

Hint:

Differentiate the function w.r.t x

Given:

$\frac{d}{d x}\left[\log \left\{e^{x}\left(\frac{x-2}{x+2}\right)^{\frac{3}{4}}\right\}\right]$

Solution:

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

$=\frac{d}{d x}\left[x \log e+\frac{3}{4} \log \left(\frac{x-2}{x+2}\right)\right]$

$y=\frac{d}{d x}\left[x+\frac{3}{4} \log \left(\frac{x-2}{x+2}\right)\right]$

$\frac{d y}{d x}=1+\frac{3}{4\left(\frac{x-2}{x+2}\right)} \times \frac{(x+2)-(x-2)}{(x+2)^{2}}$

\begin{aligned} &=1+\frac{3}{4} \frac{(x+2)}{(x-2)} \times \frac{x+2-x+2}{(x+2)^{2}} \\\\ &=1+\frac{3}{4} \frac{(x+2)}{(x-2)} \times \frac{4}{(x+2)} \end{aligned}

\begin{aligned} &=1+\frac{3}{\left(x^{2}-4\right)} \\\\ &\frac{d y}{d x}=\frac{x^{2}-4+3}{x^{2}-4}=\frac{x^{2}-1}{x^{2}-4} \end{aligned}

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