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
If y= e^x - e^-x / e^x + e^-x , then y'=
Please help
4 answers
You will need brackets
y= (e^x - e^-x) / (e^x + e^-x)
wonder what would happen if we multiply top and bottom by e^x
= y= (e^x - e^-x) / (e^x + e^-x) * (e^x/e^x)
= (e^2x - 1)/(e^2x + 1)
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
check my work, I should have written it out on paper first
y= (e^x - e^-x) / (e^x + e^-x)
wonder what would happen if we multiply top and bottom by e^x
= y= (e^x - e^-x) / (e^x + e^-x) * (e^x/e^x)
= (e^2x - 1)/(e^2x + 1)
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
check my work, I should have written it out on paper first
I know the correct answer is 4 / (e^x + e^-x)^2
You should also know that
y = tanh(x)
y' = sech^2(x) = [2/(e^x+e^-x)]^2
y = tanh(x)
y' = sech^2(x) = [2/(e^x+e^-x)]^2