Question

How do you differentiate `f(x)=A/(B+Ce^x)`?

Answer 1

`f'(x)=-(ACe^x)/(B+Ce^x)^2`

Explanation:

If we want to differentiate with the Quotient Rule:

`f'(x)=((B+Ce^x)d/dxA-Ad/dx(B+Ce^x))/(B+Ce^x)^2`

`d/dxA=0`, the derivative of a constant is zero.

`d/dx(B+Ce^x)=Ce^x`

Thus,

`f'(x)=-(ACe^x)/(B+Ce^x)^2`

Answer 2

`f'(x)=-\frac{A*C*e^x}{(B+C*e^x)^2}`

Explanation:

Remember:
`(\frac{f(x)}{g(x)})^'=\frac{f'(x)g(x)-f(x)g'(x)}{(g(x))^2}`
in this case `f(x)=A` the derivative of a constant is zero
and `g(x)=(B+C*e^x)`, it become:
`(\frac{A}{g(x)})^'=\frac{-Ag'(x)}{(g(x))^2}`
`(\frac{A}{g(x)})^'=\frac{-A(B+C*e^x)^'}{(B+C*e^x)^2}=`
`=\frac{-A*C*e^x}{(B+C*e^x)^2}`