Question

Differentiate the function with respect to x:
$\frac{x+e^x}{1+\log x}$

Answer 1

Given:
$f\left(x\right)=\frac{x+e^x}{1+\log x}$

$f^{\prime }\left(x\right)=\frac{d}{dx}\left(\frac{x+e^x}{1+\log x}\right)$

Applying quotient rule

$\left(\frac{d}{dx}\right(\frac{u}{v})=\frac{c\frac{du}{dx}-u\frac{dv}{dx}}{v^2})$ , we get

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

$\Rightarrow f^{\prime }\left(x\right)=\frac{(1+\log x)(1+e^x)-(x+e^x)\frac{1}{x}}{(1+\log x{)}^2}$

$\Rightarrow f^{\prime }\left(x\right)=\frac{x(1+\log x)\left(1_e^x\right)-(x+e^x)}{x(1+\log x{)}^2}$

$x(1+\log x+e^x+e^x\log x)$
$\Rightarrow f^{\prime }\left(x\right)=\frac{-x-e^x}{x(1+\log x{)}^2}$

$x+x\log x+x{e}^x+$
$\Rightarrow f^{\prime }\left(x\right)=\frac{x{e}^x\log x-x-e^x}{x(1+\log x{)}^2}$

$\Rightarrow f^{\prime }\left(x\right)=\frac{x\log x.\left(1_e^x\right)-e^x(1-x)}{x(1+\log x{)}^2}$

Therefore, the differentiation of the given
function is $\frac{x\log x.(1+e^x)-e^x(1-x)}{x(1+\log x{)}^2}$