Question

What is the derivative of `(e^x-e^-x)/(e^x+e^-x)`?

Answer 1

`y^' = (4e^(2x))/(e^(2x) + 1)^2`

Explanation:

You can differentiate this function by using the quotient rule and the chain rule.

Before using these rules to differentiate the function, try to simplify it a little - this will save you a lot of work on the derivative. For example, you could write

`y = (color(red)(cancel(color(black)(e^x))) * (1 - e^(-2x)))/(color(red)(cancel(color(black)(e^x))) * (1 + e^(-2x))) = (1 - 1/e^(2x)) * (1/(1 + 1/e^(2x)))`

`y = (e^(2x) - 1)/color(red)(cancel(color(black)(e^2x))) * color(red)(cancel(color(black)(e^2x)))/(e^(2x) + 1) = (e^(2x) -1)/(e^(2x) + 1)`

So, you know that you can differentiate a function that takes the form

`y = f(x)/g(x)`, with `g(x)!=0`

by using the formula

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

In your case, you have `f(x) = e^(2x)-1` and `g(x) = e^(2x)+1`, so you can write

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

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

`y^' = (color(red)(cancel(color(black)(2e^4x))) + 2 * e^(2x) - color(red)(cancel(color(black)(e^4x))) + 2 * e^(2x))/(e^(2x) + 1)^2`

`y^' = color(green)((4e^(2x))/(e^(2x) + 1)^2)`

Answer 2

`d/dx((e^x-e^(-x))/(e^x+e^(-x))) = sech^2 x`

Explanation:

Hyperbolic function:
`sinh z = 1/2(e^z - e^(-z))`
`cosh z = 1/2(e^z + e^(-z))`
`tanh z = sinhz/cosh z`
`sech z = 1/cosh z`

`d/dx((e^x-e^(-x))/(e^x+e^(-x))) = d/dx(tanh x) = sech^2 x = (2/(e^x + e^(-x)))^2 = (4e^(2x))/(e^(2x)+1)^2`