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

1 Answer
A. S. Adikesavan
Apr 5, 2016

If it is operator #(sinh)^79#, operating oh the operand x and # f_79=((sinh)^79)x#, the derivative #f'_79# is the product #(cosh f_78) (cosh f_77)(cosh f_76)...(cosh f_1)(cosh x)#

Explanation:

Let #f_79=(sinh^79)(x)#.

Then #f_79=sinh (f_78)#.

It follows that

#f'_79=cosh (f_78) f'_78#, using function pf function rule.

This is a reduction formula.

Successive reduction leads to

#f'_79=(cosh f_78) (cosh f_77)(cosh f_76)...(cosh f_1)(cosh x)#

#f_1=sinh x. f'_1=cosh x#. .

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