Question

If $f\left(x\right)=\left(\frac{lo{g}_{cotx}tanx}{lo{g}_{tanx}cotx}\right)+{\tan }^{-1}\left(\frac{4x}{\sqrt{4-x^2}}\right)$, then $f^{\prime }\left(2\right)$ is equal to

• A. 2
• B. 0
• C. $\frac{1}{2}$
• D. -2

Answer 1

Answer: (C)

$\frac{1}{2}$

$f\left(x\right)=\left(lo{g}_{cotx}tanx\right)\frac{1}{lo{g}_{tanx}cotx}+ta{n}^{-1}\left(\frac{4x}{\sqrt{4-x^2}}\right)$
$\frac{logtanx}{logcotx}\cdot \frac{logtanx}{logcotx}+ta{n}^{-1}\left(\frac{4x}{\sqrt{4-x^2}}\right)x=2sin\theta$
$\frac{lo{g}^{2}tanx}{lo{g}^{2}tanx}ta{n}^{-1}\left(\frac{8sin\theta }{4cos\theta }\right)$
$f\left(2sin\theta \right)=ta{n}^{-1}\left(2tan\theta \right)$
$f^1\left(2sin\theta \right)cos\theta =\frac{1}{1+(2tan\theta {)}^2}\left(2xe{c}^2\theta \right)$
$sin\theta =1\Rightarrow \theta =\frac{\pi }{2}$
$f^1\left(2\right)=\frac{1}{2}$