# Calculus

How to derive cosh (3x) in terms of cosh (x) and sinh (x) ? Initially I start with 1/2 (e^3x+e^-3x) but I'm stuck from there because I don't know how to continue... :(

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1. Using the chain rule:

u = 3 x

du = 3 dx Divide both sides by 3

du / 3 = dx

dx = du / 3

d cosh ( 3 x ) / dx =

d cosh ( u ) / ( du / 3 ) =

3 d cosh ( u ) / du

d cosh ( u ) / du = sinh u

so

d cosh ( 3 x ) / dx = 3 d cosh ( u ) / du = 3 sinh u = 3 sinh ( 3 x )

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2. cosh(3x) = cosh(2x+x)
= cosh2x coshx + sinh2x sinhx
...

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