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  4. If f(x)=(logcotxtanx)(logtanxcotx)-1 +tan-1 (4x/4-x2), then f'(2) is equal to

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Q. If $f\left(x\right)=\left(log_{cotx}tanx\right)\left(log_{tanx}cotx\right)^{-1}$ $+tan^{-1} \frac{4x}{4-x^{2}}$, then $f'(2)$ is equal to

Continuity and Differentiability

Solution:

$f \left(x\right)=\left(log_{cot\,x}\,tan\,x\right)\left(log_{tan\,x}\,cot\,x\right)^{-1}+tan^{-1} \frac{4x}{4-x^{2}}$
$=\frac{log\,tan\,x}{log\,cot\,x}\cdot\frac{log\,tan\,x}{log\,cot\,x}+tan^{-1}\left(\frac{4x}{4-x^{2}}\right)$
$=\frac{\left(log\,tan\,x^{2}\right)}{\left(-log\,tan\,x^{2}\right)}+tan^{-1}\left(\frac{4x}{4-x^{2}}\right)$
$=1+tan^{-1}\left(\frac{4x}{4-x^{2}}\right)$
$\therefore f '\left(x\right)=\frac{1}{1+\left(\frac{4x}{4-x^{2}}\right)^{2}}\cdot\frac{\left(4-x^{2}\right)4-4x\left(-2x\right)}{\left(4-x^{2}\right)^{2}}$
$=\frac{16-4x^{2}+8x^{2}}{\left(4-x^{2}\right)^{2}+16x^{2}}$
$=\frac{4\left(4+x^{2}\right)}{\left(4-x^{2}\right)^{2}+\left(4x\right)^{2}}$
Hence, $f '\left(2\right)=\frac{4\left(4+4\right)}{0+\left(8\right)^{2}}$
$=\frac{30}{64}=\frac{1}{2}$