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

2 Answers
Apr 16, 2018

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!

Apr 18, 2018

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!