/sci/ - Science & Math » Thread #11054862

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Anonymous Sun 13 Oct 12:57:59 2019 No.11054862 View Reply Original Report

Quoted By: >>11055057 >>11055095

If f?k (x) = f(f(f(f(f( ... x ... {k times}, state the formula for the nth derivative of f?k (x).

Anonymous

Anonymous Sun 13 Oct 2019 12:59:46 No.11054866 Report

Quoted By:

You should be able to solve this.

Anonymous

Anonymous Sun 13 Oct 2019 14:06:17 No.11054994 Report

Quoted By: >>11055013

Is f'k(x) * f'(k-1)(x) *... f'1(x) * x' the first derivative?

Anonymous

Anonymous Sun 13 Oct 2019 14:13:43 No.11055013 Report

Quoted By:

>>11054994
No.

Anonymous

Anonymous Sun 13 Oct 2019 14:24:25 No.11055044 Report

Quoted By:

There's a link to combinatorics here, but I can't find it. Anyone?

Anonymous

Anonymous Sun 13 Oct 2019 14:29:41 No.11055057 Report

Quoted By: >>11055073

>>11054862
(f(f(x)))' = f'(f(x))f'(x)
and if we write the k-th iteration as f_k then
(f_k)' = f(f_{k-1}(x))' = f'(f_{k-1}(x))f'_{k-1}(x)
expand recursively

Anonymous

Anonymous Sun 13 Oct 2019 14:34:26 No.11055073 Report

Quoted By: >>11055099

>>11055057
Helpful, but this is merely the first derivative, and I'd prefer the answer to contain fully simplified derivatives.

Anonymous

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Anonymous Sun 13 Oct 2019 14:40:23 No.11055095 Report

Quoted By:

>>11054862
If I'm not mistaken do this n times?

Anonymous

Anonymous Sun 13 Oct 2019 14:41:33 No.11055099 Report

Quoted By:

>>11055073
the first derivative is a product where the n-th term looks like f'(f_{k-n1}(x))

Just use the generalized product rule for the higher derivatives
https://en.wikipedia.org/wiki/General_Leibniz_rule#More_than_two_factors

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