What is the DERIVATIVE of y = ((e^x - e^-x) / (e^x + e^-x))?

Question

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

A) 0
B) 1
C) 2/(e^x + e^-x)²
D) 4/(e^x + e^-x)²
E) 1/(e^2x + e^-2x)

Thank you.

Answer 1

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



dy/dx

= [(e^x + e^-x)(e^x + e^-x) - (e^x - e^-x)(e^x - e^-x)]/(e^x + e^-x)²

= [(e^x + e^-x)² - (e^x - e^-x)²]/(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)²

= [2e^x][2e^-x]/((e^x + e^-x)²

= 4/(e^x + e^-x)² ==> ANSWER!!!!

Answer 2

1) u = (e^x - e^-x); ==> d(u) = (e^x + e^-x)



2) v = (e^x + e^-x); ==> d(u) = (e^x - e^-x)



3) Applying u/v form. derivative of (u/v) = [(du)*(v) - (u)*(dv)]/(v^2)



==>, dy/dx = [(e^x + e^-x)*(e^x + e^-x) - (e^x - e^-x)*(e^x - e^-x)]/(e^x + e^-x)^2



Here numerator is in he form of (a+b)^2 - (a-b)^2 = 4ab



==> [(e^x + e^-x)*(e^x + e^-x) - (e^x - e^-x)*(e^x - e^-x)] = 4(e^x)*(e^-x) = 4



Thus answer is: 4/(e^x + e^-x)^2 [Choice - D]

Answer 3

y = sinh x/cosh x



dy/dx = (cosh x*cosh x - sinh x*sinh x) / (cosh x)^2 = 1 / (cosh x)^2 = 4/(e^x + e^-x)^2