Oberlaudacu

Answered

2021-12-23

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

$y=\frac{{e}^{x}}{x};\text{}(1,{e}^{x})$

Answer & Explanation

Jenny Sheppard

Expert

2021-12-24Added 35 answers

Consider the curve

The objective is to find an equation of the tangent line to the given curve at the point

Consider the curve

Here,

In the given question.

Differentiate

Evaluate

Thus, equation of tangent line is:

Therefore, equation of the tangent line to the given curve at the point

Maricela Alarcon

Expert

2021-12-25Added 28 answers

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

Use the quotient rule,

Using the chain rule,

The derivative of x is 1:

The derivative of x is 1:

Simplify the expression:

Answer:

user_27qwe

Expert

2021-12-30Added 230 answers

First answer is right!

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