Question

Differentiate the following functions with respect to x:
$(xsinx+cosx)(e^x+x^{2}logx)$

Answer 1

We have,
$\frac{d}{dx}\left(\right(xsinx+cosx\left)\right(e^x+x^{2}logx\left)\right)$
We will apply product rule,
$=(e^x+x^{2}logx)\frac{d}{dx}(xsinx+cosx)+(xsinx+cosx)\frac{d}{dx}(e^x+x^{2}logx)$
$=(e^x+x^{2}logx)\left(\frac{d}{dx}\right(xsinx)+\frac{d}{dx}cosx)+(xsinx+cosx)\times \left\{\frac{d}{dx}\right(e^{\prime })+\frac{d}{dx}(x^{2}logx\left)\right\}$
Again apply product rule,
$=(e^x+x^{2}logx)\left(sinx\frac{d}{dx}\right(x)+x\frac{d}{dx}(sinx\left)\right)-sinx+(xsinx+cosx)\{e^x+\left(logx\frac{d}{dx}\right(x^2+x^2\frac{d}{dx}\left(logx\right)\left)\right)\}$
$=(e^x+x^{2}logx)(sinx+xcosx-sinx)+(xsinx+cosx)(e^x+logx\times 2x+x^2\frac{1}{x})$
$=(e^x+x^2)logx)\times cosx+(xsinx+cosx\left)\right(e^x+2x\times logx+x)$
$=xcosx{e}^x+e^{3}cosx+(xsinx+cosx)(e^x+2x\times logx+x)$
$=xcosx{e}^x+e^{3}cosxlogx+x{e}^{x}sinx+e^{x}cosx+2x^{2}sinx\times logx+2xcosxlogx+x^{2}sinx+xcosx$
$=xcosx(e^x+x^{2}logx)+(xsinx+cosx)(e^x+x+2xlogx)$