What is the derivative of sinh(x)sinh(x)?

1 Answer
Dec 12, 2014

(d(sinh(x)))/dx = cosh(x)d(sinh(x))dx=cosh(x)

Proof: It is helpful to note that sinh(x):=(e^x-e^-x)/2 and cosh(x):=(e^x+e^-x)/2. We can differentiate from here using either the quotient rule or the sum rule. I'll use the sum rule first:
sinh(x) = (e^x-e^-x)/2
= (e^x)/2-(e^-x)/2
=>(d(sinh(x)))/dx = d/dx((e^x)/2-(e^-x)/2)
=d/dx(e^x/2)-d/dx(e^-x/2) by the sum rule
=e^x/2-(-e^x/2) by basic differentiation of exponential functions
=(e^x+e^-x)/2 = cosh(x).

The quotient rule is just as easy:
Let u=e^x-e^-x and v=2
Hence (du)/dx=e^x+e^-x (by the sum rule and basic differentiation of exponential functions)
and (dv)/dx=0
Recalling that (d(u/v))/dx = (v((du)/dx)-u((dv)/dx))/v^2
so (d(sinh(x)))/dx=(2*(e^x+e^-x)-0)/2^2
=(2*(e^x+e^-x))/4
=(e^x+e^-x)/2=cosh(x).