Question

If $y={\left(\tan x\right)}^{cotx}$, then $\frac{dy}{dx}=$

• A. $y\left(\cos e{c}^{2}x\left(1-\log \left(\tan x\right)\right)\right)$
• B. $\frac{dy}{dx}=y\left(\cos e{c}^{2}x\left(1+\log \left(\tan x\right)\right)\right)$
• C. $\frac{dy}{dx}=y\left(\cos e{c}^{2}x\left(\log \left(\tan x\right)\right)\right)$
• D. none of these

Answer 1

The correct option is A

$y\left(\cos e{c}^{2}x\left(1-\log \left(\tan x\right)\right)\right)$

Explanation for the correct option:

Find the value of $\frac{dy}{dx}$:

Given,

$y={\left(\tan x\right)}^{cotx}$

Take $\log$ on both sides,

Then,

$$\begin{gathered} \log y=\log {\left(\tan x\right)}^{cotx} \\ \log y=cotx\log \left(\tan x\right)\left[\because log{a}^b=bloga\right] \end{gathered}$$

Now differentiate with respect to $x$.

$$\begin{gathered} \frac{1}{y}\frac{dy}{dx}=-\cos e{c}^{2}x\log \left(\tan x\right)+cotx\frac{1}{\tan x}se{c}^{2}x\left[\mathbf{\because }\mathbf{}\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathit{l}\mathit{o}\mathit{g}\mathbf{}\mathit{x}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{1}}{\mathbf{x}}\mathbf{,}\mathbf{}\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathit{t}\mathit{a}\mathit{n}\mathbf{}\mathit{x}\mathbf{}\mathbf{=}\mathit{s}\mathit{e}{\mathit{c}}^{\mathbf{2}}\mathit{x}\mathbf{}\mathbf{,}\mathbf{}\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{}\mathit{c}\mathit{o}\mathit{t}\mathbf{}\mathit{x}\mathbf{}\mathbf{=}\mathbf{-}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathit{e}{\mathit{c}}^{\mathbf{2}}\mathit{x}\mathbf{}\mathbf{}\right] \\ \Rightarrow \frac{dy}{dx}=y\left(\cos e{c}^{2}x\left(1-\log \left(\tan x\right)\right)\right) \end{gathered}$$

Hence, the correct option is A.