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.

\(y-e=0(x-1)\)

\(y=e\)

\(e^x\cdot x^{-1}\)

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.

\(y-e=0(x-1)\)

\(y=e\)