# Thread: Help with this derivative?

1. ## Help with this derivative?

It is from the infinite series section of the book, but I must have missed class this day because I don't know how to solve it. I don't need it worked out, just a point in the right direction. The problem is to compute the 7th derivative of

$1/14xcos(2x)$

I know he doesn't expect us to manually take the derivative 7 times, so what is the trick?

2. Thank you, but I understand how to manually compute the 7th derivative with the product rule. This is not what I am being tested on. It has something to do with series, but I can't find another example like this in my notes.

3. That is going to get pretty messy (this is what wolfram believes the 7th derivative is http://www.wolframalpha.com/input/?i...cos%282x%29%29 ) but anyways i believe you would use:

$cos(x) =\displaystyle\sum_{n=0}^{\infty}(-1)^n * \frac{x^{2n}}{(2n)!}$

you would pull the $\frac{1}{14x}$ out front of the summation and then plug in 2x for the x.

then you would just do n=0 to n=6 i believe to get the 7th derivative.

Hopefully someone can say if this is right or not cause i am not 100% positive....

And since your function is 1/cosx this may not work for it but its all i could think of doing from a SERIES

4. I think the series you wrote is actually sin, not cos. Can anyone clarify how to do this?

5. Yeah sorry i edited it so now it is cos. Hopefully someone smarter than me will see this post and be able to help you out a little more

is your function $\frac{1}{14}xcos(2x)$ or is it $\frac{1}{14xcos(2x)}$

if it is $\frac{1}{14}xcos(2x)$ i do think you can use this series i stated.

You would do n=o to n=7 to get 7th derivative so this is what i came up with.

for n=0 ----- $\frac{1}{14}*1$

n=1 ------ $\frac{1}{14}*(-1)*\frac{4x^2}{2!}$

n=2 ------ $\frac{1}{14}*(1)*\frac{4^4x^4}{4!}$

keep doing this until n=7 then just reduce everything and i believe you would get your 7th derivative.

6. If you are indeed correct, then I think I understand. Thank you! And it is (1/14)x cos(2x). I should have made that clear from the beginning.

7. when do you need this done by?

I can ask someone tomorrow who will know how to do it and can post it by tomorrow afternoon.

8. Test is tomorrow, but no worries.

9. oh aight well good luck on the test.

Hopefully someone will look over this post and answer the question for you.

Thanks!