What is the derivative of e^x / (e^x+1)exex+1?

1 Answer
Aug 6, 2015

y^' = e^x/(e^x + 1)^2

Explanation:

You can differentiate this function by using the quotient rule, which tells you that you can differentiate functions that take the form

f(x) = g(x)/(h(x)), with g(x)!=0

by using the formula

color(blue)(d/dx(f(x)) = ([d/dx(g(x))] * h(x) - g(x) * d/dx(h(x)))/[(g(x)]^2)

In your case, you have g(x) = e^x and h(x) = e^x + 1, which means that you can write

d/dx(f(x)) = ([d/dx(e^x)] * (e^x - 1) - e^x * d/dx(e^x - 1))/(e^x + 1)^2

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

f^' = (color(red)(cancel(color(black)(e^(2x)))) + e^x - color(red)(cancel(color(black)(e^(2x)))))/(e^x + 1)^2 = color(green)(e^x/(e^x + 1)^2)

Alternatively, you can play around with the function a bit and write it as

f(x) = e^x * (e^x + 1)^(-1)

In this case, you can the product rule and the chain rule to differentiate the function

d/dx(f(x)) = d/dx(e^x) * (e^x + 1)^(-1) + e^x * d/dx(e^x + 1)^(-1)

f^' = e^x * (e^x + 1)^(-1) + e^x * (-1) * (e^x + 1)^(-2) * e^x

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

Once again, you have

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

f^' = (color(red)(cancel(color(black)(e^(2x)))) + e^x - color(red)(cancel(color(black)(e^(2x)))))/(e^x + 1)^2 = color(green)(e^x/(e^x + 1)^2)