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  4. If f(x) = log [ex ((3-x/3+x))1/3], then f'(1) is equal to

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

KEAMKEAM 2012Continuity and Differentiability

Solution:

$\text { 67. }$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 $x=1$
$f^{\prime}(1)=1+\frac{1}{3}\left(\frac{-6}{9-1^{2}}\right)$
$=1+\frac{1}{3}\left(-\frac{6}{8}\right)=1-\frac{1}{4}=\frac{3}{4}