Question

If$f\left(x\right)=\left({\log }_{\cos }\tan x\right){({\log }_{\tan x}cotx)}^{-1}+{\tan }^{-1}\frac{x}{\sqrt{4-x^2}}$ then $f\text{'}\left(1\right)$ is

• A. $0$
• B. $-2$
• C. $\frac{1}{\sqrt{3}}$
• D. $\sqrt{3}$

Answer 1

The correct option is C

$\frac{1}{\sqrt{3}}$

Explanation for correct option:

Step 1: Simplifying the given function:

$\begin{array}{rcl}f\left(x\right)& =& \frac{\log \tan x}{\log cotx}{\left(\frac{\log cotx}{\log \tan x}\right)}^{-1}+{\tan }^{-1}\frac{x}{\sqrt{4-x^2}}\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{l}\mathbf{o}{\mathbf{g}}_{\mathbf{a}}\mathbf{b}\mathbf{=}\frac{\mathbf{l}\mathbf{o}\mathbf{g}\mathbf{b}}{\mathbf{l}\mathbf{o}\mathbf{g}\mathbf{a}}\mathbf{]}\\ & =& \frac{\log \tan x}{-\log \tan x}.\frac{\log \tan x}{-\log \tan x}+{\tan }^{-1}\frac{x}{\sqrt{4-x^2}}\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{l}\mathbf{o}\mathbf{g}\left(cotx\right)\mathbf{=}\mathbf{l}\mathbf{o}\mathbf{g}\left(\frac{1}{tanx}\right)\mathbf{=}\mathbf{-}\mathbf{l}\mathbf{o}\mathbf{g}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{x}\mathbf{]}\\ & =& 1+{\tan }^{-1}\frac{2\sin \theta }{\sqrt{4-4{\sin }^2\theta }}\left[\mathrm{putting}\mathrm{x}=2\mathrm{sin\theta }\right]\\ & =& 1+{\tan }^{-1}\frac{2\sin \theta }{2\cos \theta }\\ & =& 1+{\tan }^{-1}\left(\tan \theta \right)\\ & =& 1+\theta \\ & =& 1+{\sin }^{-1}\frac{x}{2}\left[\mathrm{x}=2\mathrm{sin\theta }\right]\end{array}$

Step 2: Find the value of $f\text{'}\left(1\right)$:

Differentiated the simplified function with respect to $x$.

$$\begin{gathered} f\text{'}\left(x\right)=\frac{1}{\sqrt{1-{\displaystyle \frac{x^2}{4}}}}.\frac{1}{2} \\ =\frac{1}{\sqrt{4-x^2}} \end{gathered}$$

Substitute $x=1$ in $f\text{'}\left(x\right)$, we get

$\begin{array}{rcl}f\text{'}\left(1\right)& =& \frac{1}{\sqrt{4-1}}\\ & =& \frac{1}{\sqrt{3}}\end{array}$

Hence, the correct option is C.