Value of nth derivative using Malcurian Series

**Planeman** · December 14th 2009, 04:33 PM

I have a question:
Given: $f(x)=x^7 \cos(x^9)$

Find $f^{79}(0)$ , the 79-th derivative of f(x) at x=0

I know the answer is $\frac{79!}{8!}$ but I am not sure of all the steps.

I found the Taylor Series representation to be:
$\sum\frac{(-1)^nx^{18n+7}}{2n!}$

Any enlightenment would be much appreciated.

**Drexel28** · December 14th 2009, 07:26 PM

Originally Posted by Planeman

I have a question:
Given: $f(x)=x^7 \cos(x^9)$

Find $f^{79}(0)$ , the 79-th derivative of f(x) at x=0

I know the answer is $\frac{79!}{8!}$ but I am not sure of all the steps.

I found the Taylor Series representation to be:
$\sum\frac{(-1)^nx^{18n+7}}{2n!}$

Any enlightenment would be much appreciated.

We know that $\frac{f^{79}(0)}{n!}$ is the coefficient of the term containing $x^{79}$ term in the power series. Noting though that $18n+7=79\implies n=4$ . Thus $\frac{f^{79}(0)}{79!}=\frac{(-1)^4}{(2\cdot4)!}\implies f^{79}(0)=\frac{79!}{8!}$ .

**Planeman** · December 14th 2009, 07:44 PM

Thank you so much, now it all makes sense.

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