Question

How do you differentiate `f(x) =1/(e^(3-x)+1)` using the quotient rule?

Answer 1

`f'(x)=(e^(x+3))/(e^x+e^3)^2`

Explanation:

According to the quotient rule,

`f'(x)=((e^(3-x)+1)d/dx(1)-1d/dx(e^(3-x)+1))/(e^(3-x)+1)^2`

Finding both of the internal derivatives, we see that

`d/dx(1)=0`

To find the second derivative, use the chain rule:

`d/dx(e^x)=e^x" "=>" "d/dx(e^u)=e^u*(du)/dx`

Since `u=3-x`,

`d/dx(e^(3-x)+1)=e^(3-x)d/dx(3-x)=e^(3-x)(-1)=-e^(3-x)`

The `1` will be neglected since it is a constant.

Plug these both back in:

`f'(x)=((e^(3-x)+1)(0)-1(-e^(3-x)))/(e^(3-x)+1)^2`

`f'(x)=e^(3-x)/(e^(3-x)+1)^2`

We can try to simplify further:

`f'(x)=(e^(3-x))/(e^3/e^x+e^x/e^x)^2`

`f'(x)=(e^(3-x))/(((e^3+e^x)^2)/e^(2x))`

`f'(x)=(e^(3-x)e^(2x))/(e^x+e^3)^2`

`f'(x)=(e^(x+3))/(e^x+e^3)^2`