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

Please help i know how to get the derivative i just can not simplify.

To find the derivative of y = (e^x - e^-x)/(e^x + e^-x), we can use the quotient rule.

The quotient rule states that for two functions u(x) and v(x), where u(x) = f(x)/g(x), the derivative of u(x) with respect to x is given by:

(u/v)' = (u'v - uv')/v^2

Applying the quotient rule to the given function y = (e^x - e^-x)/(e^x + e^-x), we have:

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

Simplifying this expression will allow us to find y'.

First, let's expand the numerator and the denominator separately:

Numerator:
(e^x + e^-x)(e^x + e^-x) = e^x * e^x + e^x * e^-x + e^-x * e^x + e^-x * e^-x
= e^2x + 2 * e^x * e^-x + e^-2x

Denominator:
(e^x + e^-x)^2 = (e^x + e^-x)(e^x + e^-x) = e^x * e^x + e^x * e^-x + e^-x * e^x + e^-x * e^-x
= e^2x + 2 * e^x * e^-x + e^-2x

Now, substitute these expanded expressions back into the numerator and denominator in the original expression for y'.

Numerator:
(e^2x + 2 * e^x * e^-x + e^-2x) - (e^2x - 2 * e^x * e^-x + e^-2x)
= e^2x + 2 * e^x * e^-x + e^-2x - e^2x + 2 * e^x * e^-x - e^-2x
= 4 * e^x * e^-x = 4

Denominator:
e^2x + 2 * e^x * e^-x + e^-2x

Therefore, y' = 4/(e^2x + 2 * e^x * e^-x + e^-2x).

To summarize, the derivative of y = (e^x - e^-x)/(e^x + e^-x) is y' = 4/(e^2x + 2 * e^x * e^-x + e^-2x).