Question

If $f\left(x\right)=log\left[e^x{\left(\frac{3-x}{3+x}\right)}^{1/3}\right]$, then $f^{\prime }\left(1\right)$ is equal to

• A. $\frac{3}{4}$
• B. $\frac{2}{3}$
• C. $\frac{1}{3}$
• D. $\frac{1}{2}$
• E. $\frac{1}{4}$

Answer 1

Answer: (A)

$\text{ 67. }$Given,$f(x)=\log \left[e^x{\left(\frac{3-x}{3+x}\right)}^{1/3}\right]$
$\Rightarrow f(x)=\log e^x+\frac{1}{3}[\log (3-x)-\log (3+x)]$
On differentiating w.r.t. $x$, we get
$f^{\prime }(x)=\frac{1}{e^x}\times e^x+\frac{1}{3}\left[\frac{(-1)}{3-x}-\frac{1}{3+x}\right]$
$=1+\frac{1}{3}\left[-\frac{1}{3-x}-\frac{1}{3+x}\right]$
$=1+\frac{1}{3}\left[\frac{-3-x-3+x}{9-x^2}\right]$
$=1+\frac{1}{3}\left(\frac{-6}{9-x^2}\right)$
$At$x=1$<br/><br/>$f^{\prime}(1)=1+\frac{1}{3}\left(\frac{-6}{9-1^{2}}\right)$<br/><br/>$=1+\frac{1}{3}\left(-\frac{6}{8}\right)=1-\frac{1}{4}=\frac{3}{4}