Question

What is the equation of the tangent line of `f(x) =xlnx-e^(3/x)` at `x = 2`?

Answer 1

See below

Explanation:

We look for equation line `y-y_0=m(x-x_0)` where `m` is the slope and `(x_0,y_0)` is the passing point.

The coordinates of passing point is `(2,f(2))=(2,2ln2-e^(3/2))`

We know that derivative valuate in `x_0` gives the slope of tangent line, thus

`f´(x)=1·lnx+x·1/x-e^(3/x)·(-3/x^2)=lnx+1+(3e^(3/x))/x^2`

In `x_0=2` we have `f´(2)=ln2+1+3/4e^(3/2)=m`

Then our line has equation

`y-2ln2+e^(3/2)=(ln2+1+3/4e^(3/2))(x-2)`