Question

The derivative of $(x^2+1)\log x-e^x\cos xw.r.tx$ is

• A. $2x\log x-x+1/x-e^x(\sin x-\cos x)$
• B. $2x\log x+x+1/x+e^x(\sin x-\cos x)$
• C. $2x\log x+x-1/x+e^x(\sin x-\cos x)$
• D. $2x\log x-x+1/x+e^x(\sin x-\cos x)$

Answer 1

The correct option is B $2x\log x+x+1/x+e^x(\sin x-\cos x)$

Let $y=(x^2+1)\log x-e^x\cos x$

Differentiate both sides w.r.t. $x,$ we get :

$\Rightarrow$ $\frac{dy}{dx}=(x^2+1)\frac{d}{dx}\left(\log x\right)+\log x\frac{d}{dx}(x^2+1)-\left(e^x\frac{d}{dx}\right(\cos x)+\cos x\frac{d}{dx}(e^x\left)\right)$

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

$\Rightarrow$ $\frac{dy}{dx}=x+\frac{1}{x}+2x\log x+e^x(\sin x-\cos x)$

$2x\log x+x+1/x+e^x(\sin x-\cos x)$