Question

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

Answer 1

Equation of tangent is `4e^3x-(3-e^3)^2y-2e^6-6e^3=0`

Explanation:

We are seeking a tangent to the curve `f(x)=(2e^x)/(x-e^x)` at `x=3` i.e. at the point `(3,f(3))` or `(3,(2e^3)/(3-e^3))`

As the slope of tangent is given by the value of its first derivative at that point, the slope is `f'(3)`.

As `f(x)=(2e^x)/(x-e^x)`, `f'(x)=((x-e^x)xx2e^x-2e^x(1-e^x))/(x-e^x)^2`

= `(2xe^x-2e^(2x)-2e^x+2e^(2x))/(x-e^x)^2`

= `(2xe^x-2e^x)/(x-e^x)^2`

= `(2e^x(x-1))/(x-e^x)^2`

and at `x=3`, slope is `(4e^3)/(3-e^3)^2`

and hence equation of tangent is

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

or `(3-e^3)^2y-2e^3(3-e^3)=4e^3x-12e^3`

or `4e^3x-(3-e^3)^2y+2e^3(3-e^3)-12e^3=0`

or `4e^3x-(3-e^3)^2y-2e^6-6e^3=0`