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

##### 1 Answer

Feb 12, 2016

#### 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

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

The

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#