Question

What is the derivative of `f(x) = x^2/(e^x-1)` at `x=0` ?

Answer 1

`1`

Explanation:

Note

Note that `color(blue)(0)^2/(e^(color(blue)(0))-1)` is actually undefined. So strictly speaking the derivative is undefined at `x=0`, since the function is.

The discontinuity is removable however, by defining the function to be zero at `x=0`, so:

`f(x) = { (x^2/(e^x-1) " if " x != 0), (0color(white)(.000) " if " x = 0) :}`

I will assume this definition below.

Quick and dirty method

`e^x - 1 = 1+x+O(x^2) - 1 = x+O(x^2)`

So:

`x^2/(e^x-1) = x^2/(x+O(x^2)) ~~ x^2/x = x`

So the value of the derivative at `0` is that of `x` at `0`, namely `1`.

graph{x^2/(e^x-1) [-5, 5, -2.5, 2.5]}

Answer 2

`1`

Explanation:

Here's a slightly different way of looking at it...

Given:

`f(x) = { (x^2/(e^x-1) " if " x != 0), ( 0color(white)(.000) " if " x = 0) :}`

The quotient rule tells us that:

`d/(dx) g(x)/(h(x)) = (g'(x)h(x)-g(x)h'(x))/(h(x))^2`

So:

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

Note that:

`e^x = sum_(k=0)^oo x^k/(k!)`

which we can approximate for small values of `x` by the quadratic function:

`e_2(x) = 1+x+1/2x^2`

Then:

`2x(e_2(x)-1)-x^2e_2(x) = 2x((1+x+1/2x^2) - 1)-x^2(1+x+1/2x^2)`

`color(white)(2x(e_2(x)-1)-x^2e_2(x)) = 2x(x+1/2x^2)-x^2(1+x+1/2x^2)`

`color(white)(2x(e_2(x)-1)-x^2e_2(x)) = 2x^2+x^3-(x^2+x^3+1/2x^4)`

`color(white)(2x(e_2(x)-1)-x^2e_2(x)) = x^2+1/2x^4`

`color(white)(2x(e_2(x)-1)-x^2e_2(x)) = x^2(1+1/2x^2)`

and:

`(e_2(x)-1)^2 = ((1+x+1/2x^2)-1)^2`

`color(white)((e_2(x)-1)^2) = (x+1/2x^2)^2`

`color(white)((e_2(x)-1)^2) = x^2(1+1/2x)^2`

`color(white)((e_2(x)-1)^2) = x^2(1+x+1/4x^2)`

Then:

`lim_(x->0) (2x(e^x-1)-x^2e^x)/(e^x-1)^2 = lim_(x->0) (2x(e_2(x)-1)-x^2e_2(x))/(e_2(x)-1)^2`

`color(white)(lim_(x->0) (2x(e^x-1)-x^2e^x)/(e^x-1)^2) = lim_(x->0) (color(red)(cancel(color(black)(x^2)))(1+1/2x^2))/(color(red)(cancel(color(black)(x^2)))(1+x+1/4x^2))`

`color(white)(lim_(x->0) (2x(e^x-1)-x^2e^x)/(e^x-1)^2) = lim_(x->0) (1+1/2x^2)/(1+x+1/4x^2)`

`color(white)(lim_(x->0) (2x(e^x-1)-x^2e^x)/(e^x-1)^2) = 1/1`

`color(white)(lim_(x->0) (2x(e^x-1)-x^2e^x)/(e^x-1)^2) = 1`

Here are the graphs of `x^2/(e^x-1)` and `x^2/(e_2(x)-1)` near `x=0`

graph{(y-x^2/(e^x-1))(y-x^2/(x+x^2/2)) = 0 [-4.79, 5.21, -2.6, 2.4]}