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