Find an equation of the tangent line to the given curve at thespecified point y=exx;

Consider the curve ${\displaystyle y=\frac{e^x}{x}}$
The objective is to find an equation of the tangent line to the given curve at the point ${\displaystyle (1,e^x)}$
Consider the curve ${\displaystyle y=f\left(x\right)}$. Then, equation of tangent at the point ${\displaystyle (x_1,y_1)}$ is given by, ${\displaystyle y-y_1=m(x-x_1)}$
Here, ${\displaystyle m=\frac{dy}{dx}{\mid }_{(x_1,y_1)}}$ =slope of the tangent at the point ${\displaystyle (x_1,y_1)}$.
In the given question. ${\displaystyle f\left(x\right)=\frac{e^x}{x}}$ and ${\displaystyle (x_1,y_1)=(1,e^x)}$.
Differentiate ${\displaystyle y=\frac{e^x}{x}}$ with respect to x:
${\displaystyle \frac{dy}{dx}=\frac{d}{dx}\left(\frac{e^x}{x}\right)}$
${\displaystyle =\frac{x\frac{d}{dx}\left(e^x\right)-e^x\frac{d}{dx}\left(x\right)}{x^2}}$
${\displaystyle =\frac{x{e}^x-e^x\left(1\right)}{x^2}}$
${\displaystyle =\frac{e^x(x-1)}{x^2}}$
Evaluate ${\displaystyle \frac{dy}{dx}}$ at ${\displaystyle (x_1,y_1)=(1,e^x)}$:
$m=\frac{d{y}_1}{d{x}_1}=\frac{e^1(1-1)}{1^2}=0$
Thus, equation of tangent line is:
${\displaystyle y-y_1=m(x-x_1)}$
${\displaystyle y-e^x=0(x-1)}$
${\displaystyle y-e^1=0}$ (at ${\displaystyle x=1}$)
${\displaystyle y=e}$
Therefore, equation of the tangent line to the given curve at the point ${\displaystyle (1,e^x)}$ is ${\displaystyle y=e}$

Possible derivation:
${\displaystyle \frac{d}{dx}\left(\frac{e^x}{x}\right)}$
Use the quotient rule, ${\displaystyle \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}}$, where ${\displaystyle u=e^x}$ and ${\displaystyle v=x}$:
${\displaystyle =\frac{x\left(\frac{d}{dx}\left(e^x\right)\right)-e^x\left(\frac{d}{dx}\left(x\right)\right)}{x^2}}$
Using the chain rule, ${\displaystyle \frac{d}{dx}\left(e^x\right)=\frac{d{e}^u}{du}\frac{du}{dx}}$, where ${\displaystyle u=x}$ and ${\displaystyle \frac{d}{du}\left(e^u\right)=e^u}$:
${\displaystyle =\frac{-e^x\left(\frac{d}{dx}\left(x\right)\right)+e^x\left(\frac{d}{dx}\left(x\right)\right)x}{x^2}}$
The derivative of x is 1:
${\displaystyle =\frac{-e^x\left(\frac{d}{dx}\left(x\right)\right)+1e^{x}x}{x^2}}$
The derivative of x is 1:
${\displaystyle =\frac{e^{x}x-1e^x}{x^2}}$
Simplify the expression:
Answer: ${\displaystyle =\frac{e^x(-1+x)}{x^2}}$

First answer is right!