Question

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

Answer 1

`(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).`