# How do you find the derivative of sinh^(79x)?

Apr 5, 2016

If it is operator ${\left(\sinh\right)}^{79}$, operating oh the operand x and ${f}_{79} = \left({\left(\sinh\right)}^{79}\right) x$, the derivative $f {'}_{79}$ is the product $\left(\cosh {f}_{78}\right) \left(\cosh {f}_{77}\right) \left(\cosh {f}_{76}\right) \ldots \left(\cosh {f}_{1}\right) \left(\cosh x\right)$

#### Explanation:

Let ${f}_{79} = \left({\sinh}^{79}\right) \left(x\right)$.

Then ${f}_{79} = \sinh \left({f}_{78}\right)$.

It follows that

$f {'}_{79} = \cosh \left({f}_{78}\right) f {'}_{78}$, using function pf function rule.

This is a reduction formula.

$f {'}_{79} = \left(\cosh {f}_{78}\right) \left(\cosh {f}_{77}\right) \left(\cosh {f}_{76}\right) \ldots \left(\cosh {f}_{1}\right) \left(\cosh x\right)$
f_1=sinh x. f'_1=cosh x. .