Question

If y= (e^x-e^-x)/(e^x+e^-x), then y'=



I took the derivative and got:

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


I know the correct answer is 4/(e^x+e^-x)^2

Answer 1

I did this for you a few days back, when you posted it as Lynds

http://www.jiskha.com/display.cgi?id=1420764439

and I just noticed an error with signs
I had:
now dy/dx
= ( (e^2x + 1)(2e^2x) - (e^2x - 1)(2e^2x) )/(e^2x + 1)^2
= (2e^4x + 2e^2x - 2e^2x + 2e^4x)/(2e^2x + 1)^2
= 4e^4x/(2e^2x + 1)^2

from the 2nd last line on it should have been

= (2e^4x + 2e^2x - 2e^4x + 2e^2x)/(2e^2x + 1)^2
= 4e^2x/(2e^2x + 1)^2

This result is verified by Wolfram:
http://www.wolframalpha.com/input/?i=derive+y%3D+%28e%5Ex-e%5E-x%29%2F%28e%5Ex%2Be%5E-x%29
If you had looked at my solution like I had suggested you would have easily spotted that erro

The fact that my answer of 4e^2x/(2e^2x + 1)^2
is the same as 4/(e^x + e^-x)^2 can be shown easily.

You also ignored Steve's comment using the definition of
tanh , which would pop out your answer directly.