Question

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

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

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.
$$y-e=0(x-1)$$
$$y=e$$