Question

Hi, can i ask what is the differentiate of `(e^x + e^-x) / (e^x - e^-x)`? Please Thanks much

Answer 1

The derivative is `-(2/(e^x- e^(-x)))^2`

Explanation:

Notice that `sinhx = (e^x - e^(-x))/2` and `coshx = (e^x+ e^(-x))/2`.

Then we can see that

`coshx/sinhx = cothx = ((e^x + e^(-x))/2)/((e^x - e^(-x))/2) = (e^x + e^(-x))/(e^x - e^(-x))`

Therefore, all we must do is differentiate `cothx`. The nice thing about hyperbolic trig functions is that they differentiate the same way as regular trigonometric functions.

Recall that `d/dx(cotx) = -csc^2x`, therefore, `d/dx(cothx) = -csc^2hx`

You can then recall that `cschx= 2/(e^x- e^(-x))` meaning that `d/dx((e^x +e^(-x))/(e^x - e^(-x))) = -(2/(e^x- e^(-x)))^2`

Hopefully this helps!

Answer 2

Here's how you would answer by the quotient rule:

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

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

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

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

As we got with the other method.

Hopefully this helps!