Oberlaudacu

2021-12-23

Find an equation of the tangent line to the given curve at thespecified point

Jenny Sheppard

Expert

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

Maricela Alarcon

Expert

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

user_27qwe

Expert