# hyperbolic functions!!

• Mar 11th 2008, 09:18 AM
mathlete
hyperbolic functions!!
Find the derivative of the function g(t) = cosh^3(t).

ok, so i know the correct answer for this one:

3cosh^2(t)sinh(t)

now i would like someone to show me how exactly to get that answer...
thanks

mathlete
• Mar 11th 2008, 09:44 AM
wingless
Let $\displaystyle h(t) = \cosh^3 t$

We can write it as,
$\displaystyle f(g(t))$ where $\displaystyle f(t) = t^3$ and $\displaystyle g(t) = \cosh t$.

Simply use the chain rule.

Chain rule:
$\displaystyle [f(g(t))]' = f'(g(t))\cdot g'(t)$
• Mar 11th 2008, 10:29 AM
mathlete
Quote:

Originally Posted by wingless
Let $\displaystyle h(t) = \cosh^3 t$

We can write it as,
$\displaystyle f(g(t))$ where $\displaystyle f(t) = t^3$ and $\displaystyle g(t) = \cosh t$.

Simply use the chain rule.

Chain rule:
$\displaystyle [f(g(t))]' = f'(g(t))\cdot g'(t)$

oh ok, so the derivative of cosh(t) is just sinh(t), thats what i was having trouble with, but i guess its easier than i was making it...
thanks