Question

Differentiate the following functions with respect to x :

$\frac{x+e^x}{1+logx}$

Answer 1

By Using quotient rule, we have

$\frac{d}{dx}\left(\frac{x+e^x}{1+logx}\right)$

$=\frac{(1+logx)\frac{d}{dx}(x+e^x)-(x+e^x)\frac{d}{dx}(1+logx)}{(1+logx{)}^2}=\frac{(1+logx)(1+e^x)-(x+e^x)\times \frac{d}{dx}}{(1+logx{)}^2}$

$=\frac{x(1+logx+e^x+e^{x}logx)-x-e^x}{x(1+logx{)}^2}=\frac{x+xlogx+x{e}^x+x{e}^{x}logx-x-e^x}{x(1+logx{)}^2}$

$=\frac{xlogx(1+e^x)-e^x(1-x)}{x(1+logx{)}^2}$