For $x^2-4
eq 0$, the value of $\frac{d}{d x}\left[\log \left\{e^x\left(\frac{x-2}{x+2} ight)^{\frac{3}{4}} ight\} ight]$ at $x=3$ is

Question

For $x^2-4 
eq 0$, the value of $\frac{d}{d x}\left[\log \left\{e^x\left(\frac{x-2}{x+2}ight)^{\frac{3}{4}}ight\}ight]$ at $x=3$ is

Aicharya · Accepted Answer

Let $y=\log \left\{e^x\left(\frac{x-2}{x+2}ight)^{\frac{3}{4}}ight\}$ $ \begin{aligned} & =\log e^x+\log \left(\frac{x-2}{x+2}ight)^{\frac{3}{4}}=x \log e+\frac{3}{4} \log \left(\frac{x-2}{x+2}ight) \ & =x+\frac{3}{4} \log \left(\frac{x-2}{x+2}ight) \end{aligned} $ Now, differentiating w.r.t. $x$, we get $ \begin{aligned} & \frac{d y}{d x}=1+\frac{3}{4} \frac{d}{d x} \log \left(\frac{x-2}{x+2}ight) \ & =1+\frac{3}{4}\left[\frac{x+2}{x-2} \cdot\left\{\frac{(x+2) \cdot 1-(x-2) \cdot 1}{(x+2)^2}ight\}ight] \ & =1+\frac{3}{4} 	imes \frac{4}{x^2-4}=1+\frac{3}{x^2-4} \ & 	ext { At } x=3, \ & \left(\frac{d y}{d x}ight)_{x=3}=1+\frac{3}{5}=\frac{8}{5} \end{aligned} $

For $x^2-4 \neq 0$, the value of $\frac{d}{d x}\left[\log…

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