# 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.

*print*Print*list*Cite

### 3 Answers

f= e^(x^2)/x ==> f(1)= e

We ned to prove that the tangent line is y=x*r at x=1

The tangent line for f is f' at x=1

f = u/v ==> f' = [u'v-uv']/v^2

u = e^(x^2) ==> u'= 2x*e^(x^2)

v= x ==> v' =1

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

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

= e^(x^2)[2x^2-1)/x^2.

Now substitute with x 1

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

Now the tangent line equation is:

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

y-e= e(x-1)

==> y = ex-e+e = ex

==> y = ex

The tangent line at (x1 y1) is given by y-y1 = {(dy/dx) at x=x1, y=y1} (x-x1)......(1)

f(x) = e^(x^2/x.(given). So f(1) = y1 = e^1/1 = e.

f'(x) = e^x^2 (-1/x^2) + (e^x^2)(2x)/x = e^x^2{-1/x^2+2}.

f'(1) = e{-1+2) = e.

Therefore, the tangent at (1,e) is

y -e = e(x-1). Or

y-e = ex -e. Or

y = ex is the tangent line at x=1

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.**