Find an equation of the tangent line to the given curve at the specified point. y=\frac{e^x}{x},(1,e)

iohanetc

iohanetc

Answered question

2021-05-31

Find an equation of the tangent line to the given curve at the specified point.
y=exx,(1,e)

Answer & Explanation

Szeteib

Szeteib

Skilled2021-06-01Added 102 answers

We can rewrite the original equation in the form
exx1
Now, we can differentiate using the Product Rule
(ex)(x1)+(ex)(x1)
ex(x1)+ex(x2)
Simplify
exxexx2
In order to find the tangent at our point (1,e). We need to plug in 1 into our deriative.
e11e212=ee=0
So 0 is our slope at point (1,e), to find the normal line we take the negative reciprocal of our slope.
10=undefined. Which states that our normal to the point (1,e) is equal to our x value, therefore the normal line to our dunction is simplify x=1
In order to find the tangent we need to use the Point-Slope formula.
ye=0(x1)
y=e

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?