Question

If $y={\left(sinx\right)}^{\tan x}$, then $\frac{dy}{dx}=$ (IIT-JEE, 1994)

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

Answer 1

Answer: (A)

${\left(\sin x\right)}^{\tan x}(1+{\sec }^{2}x\log \sin x)$

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

$\Rightarrow \log y=\tan x\log \sin x$

Differentiating w.r.t. $x$, we get

$\frac{1}{y}\cdot \frac{dy}{dx}={\sec }^{2}x\log \sin x+\tan x\cdot \frac{1}{\sin x}.\cos x$

$\Rightarrow \frac{dy}{dx}={\left(\sin x\right)}^{\tan x}[1+{\sec }^{2}x\log \sin x]$