Question

If $y={(\sin x)}^{\tan x}$, then $\frac{dy}{dx}$is equal to

• A. $\sin x^{\tan x}(1+se{c}^{2}x\log \sin x)$
• B. $\tan x{(\sin x)}^{\tan x-1}\cos x$
• C. $\sin x^{\cos x}se{c}^{2}x\log \sin x$
• D. $\tan x{(\sin x)}^{\tan x-1}$

Answer 1

The correct option is A

$\sin x^{\tan x}(1+se{c}^{2}x\log \sin x)$

Explanation for the correct option:

$y={(\sin x)}^{\tan x}$

Taking log both sides:

$$\begin{gathered} \log y=\log \left[{\left(\sin x\right)}^{\tan x}\right] \\ \Rightarrow \log y=\tan x\log (\sin x) \end{gathered}$$

Using Product Rule, Differentiating with respect to $x$ both sides:

$\begin{array}{rcl}\frac{1}{y}\frac{dy}{dx}& =& se{c}^{2}x\times \log (\sin x)+\tan x\times \frac{1}{\sin x}\times \cos x\\ & \Rightarrow & \frac{1}{y}\frac{dy}{dx}=se{c}^{2}x\log (\sin x)+\tan x\times cotx\\ & \Rightarrow & \frac{1}{y}\frac{dy}{dx}=se{c}^{2}x\log (\sin x)+1\\ & \Rightarrow & \frac{dy}{dx}=y(1+se{c}^{2}x\log (\sin x\left)\right)\\ & \Rightarrow & \frac{dy}{dx}=\sin x^{\tan x}(1+se{c}^{2}x\log \sin x)\end{array}$

Hence, Option (A) is the correct answer.