# Value of nth derivative using Malcurian Series

• December 14th 2009, 04:33 PM
Planeman
Value of nth derivative using Malcurian Series
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.
• December 14th 2009, 07:26 PM
Drexel28
Quote:

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 $\frac{f^{79}(0)}{79!}=\frac{(-1)^4}{(2\cdot4)!}\implies f^{79}(0)=\frac{79!}{8!}$.
• December 14th 2009, 07:44 PM
Planeman
Thank you so much, now it all makes sense. (Smile)