Question

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

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asked 2021-05-31
Find an equation of the tangent line to the given curve at the specified point.
\(y=\frac{e^x}{x},(1,e)\)

Answers (1)

2021-06-01
We can rewrite the original equation in the form
\(e^x\cdot x^{-1}\)
Now, we can differentiate using the Product Rule
\((e^x)'(x^{-1})+(e^x)(x^{-1})'\)
\(e^x(x^{-1})+e^x(-x^{-2})\)
Simplify
\(\frac{e^x}{x}-\frac{e^x}{x^2}\)
In order to find the tangent at our point (1,e). We need to plug in 1 into our deriative.
\(\frac{e^1}{1}-\frac{e^2}{1^2}=e-e=0\)
So 0 is our slope at point (1,e), to find the normal line we take the negative reciprocal of our slope.
\(\frac{-1}{0}=\text{undefined}\). 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.
\(y-e=0(x-1)\)
\(y=e\)
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