Question

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

(a) $\frac{x^2-1}{x^2-4}$

(b) 1

(c) $\frac{x^2+1}{x^2-4}$

(d) $e^x\frac{x^2-1}{x^2-4}$

Answer 1

(a) $\frac{x^2-1}{x^2-4}$

$$\begin{gathered} \mathrm{Let}y=\frac{d}{dx}\left[\log \left\{e^x{\left(\frac{x-2}{x+2}\right)}^{\frac{3}{4}}\right\}\right] \\ \Rightarrow y=\frac{d}{dx}\left[x\log e+\frac{3}{4}\log \left(\frac{x-2}{x+2}\right)\right] \\ \Rightarrow y=\frac{d}{dx}\left[x+\frac{3}{4}\log \left(\frac{x-2}{x+2}\right)\right] \\ \Rightarrow \frac{dy}{dx}=1+\frac{3}{4\left(\frac{x-2}{x+2}\right)}\times \frac{\left(x+2\right)\times 1-\left(x-2\right)\times 1}{{\left(x+2\right)}^2} \\ \Rightarrow \frac{dy}{dx}=1+\frac{3\left(x+2\right)}{4\left(x-2\right)}\times \frac{x+2-x+2}{{\left(x+2\right)}^2} \\ \Rightarrow \frac{dy}{dx}=1+\frac{3\left(x+2\right)}{4\left(x-2\right)}\times \frac{4}{\left(x+2\right)} \\ \Rightarrow \frac{dy}{dx}=1+\frac{3}{\left(x^2-4\right)} \\ \Rightarrow \frac{dy}{dx}=\frac{x^2-4+3}{x^2-4} \\ \Rightarrow \frac{dy}{dx}=\frac{x^2-1}{x^2-4} \end{gathered}$$