Prove that the equation of the tangent line to the graph of function f=e^(x^2)/x, in the point x=1, is y=x*e.

First, let's write the equation of the tangent line to the graph, to see what elements we have and what we have to find out.

y - f(1) = f'(1)*(x-1)

It's obvious that we have to calculate f(1), f'(x) and f'(1).

Let's begin with f(1).

We'll substitute x by 1, in the expression of the fucntion f(x).

f(1)=e^1/1

f(1)=e

Now, we'll calculate the first derivative of the function, using the product rule:

f'(x) = {[e^(x^2)]'*x - e^(x^2)*x'}/x^2

f'(x) = [2x*e^(x^2)*x - e^(x^2)]/x^2

We'll factorize and we'll get:

f'(x) = [e^(x^2)(2x^2 - 1)]/x^2

Now, we'll calculate f'(1).

f'(1) = [e^(1^2)(2 - 1)]/1^2

f'(1) = e

Now, we'll substitute the values for f(1) and f'(1), into equation of the tangent line, to verify if it's expression is the same with the one given into enunciation.

y - f(1) = f'(1)*(x-1)

y - e = e*(x-1)

We'll open the brackets and we'll get:

y - e =e*x - e

We'll reduce the similar terms:

y=e*x   q.e.d.