Question

Differentiate with respect to $x$.

$x^{\cos x}+\sin x^{\tan x}$

• A. $x^{\cos x}[\frac{\cos x}{x}-\sin x]+\sin x^{\tan x}[1+{\sec }^{2}x.\log \sin x]$
• B. $x^{\cos x}[\frac{\cos x}{x}-\sin x.\log x]+\sin x^{\tan x}[1+{\sec }^{2}x.\log \sin x]$
• C. $x^{\cos x}[\cos x-\sin x.\log x]+\sin x^{\tan x}[1+\sec x.\log \sin x]$
• D. None

Answer 1

The correct option is B $x^{\cos x}[\frac{\cos x}{x}-\sin x.\log x]+\sin x^{\tan x}[1+{\sec }^{2}x.\log \sin x]$

let $y=x^{\cos x}+\sin x^{\tan x}$

let $h=x^{\cos x}$ $\frac{d}{dx}\left(f\right(x\left)g\right(x\left)\right)=f^{\prime }\left(x\right)g\left(x\right)+f^{\prime }\left(x\right)f\left(x\right)$

applying $log$ on both sides we get

$\log h=\cos x\log x$

differentiate on both sides wrt $x$

$\frac{1}{h}\frac{dh}{dx}=-\sin x\log x+\frac{\cos x}{x}$

$\therefore \frac{dh}{dx}=x^{\cos x}(\frac{\cos x}{x}-\sin x\log x)$

let $t=\sin x^{\tan x}$

applying $\log$ on both sides

$\frac{1}{t}\frac{dt}{dx}={\sec }^{2}x\log \left(\sin x\right)+\frac{\tan x}{\sin x}\left(\cos x\right)$

$\therefore \frac{dt}{dx}=\sin x^{\tan x}(1+{\sec }^{2}x\log (\sin x\left)\right)$

$y=h+t$

$⟹\frac{dy}{dx}=\frac{dh}{dx}+\frac{dt}{dx}$

$\therefore \frac{dy}{dx}=x^{\cos x}(\frac{\cos x}{x}-\sin x\log x)+\sin x^{\tan x}(1+{\sec }^{2}x\log (\sin x\left)\right)$