# Thread: Maclaurin Series - find nth derivative

1. ## Maclaurin Series - find nth derivative

I have a WebWork problem that gives me :

f(x)=[cos(2x^2)-1]/(x^3)

and it asks me to find the 9th derivative at 0. It gives me a hint that I should use the Maclaurin series of f(x).

So far, I found f(x) in summation form. I first found cos(2x^2), subtracted -1, and then divided each term by x^3. At the end I get the equation attached to this post. I don't know how to proceed...

I thought that f(9nthDerivative)(0)/9! is equal to the constants of the ninth term in the summation. i'm really stuck. Can someone help me?

Ps. sorry, I'm new to the forum and I don't know how to add equations :S

2. Originally Posted by makarooney
I have a WebWork problem that gives me :

f(x)=[cos(2x^2)-1]/(x^3)

and it asks me to find the 9th derivative at 0. It gives me a hint that I should use the Maclaurin series of f(x).

So far, I found f(x) in summation form. I first found cos(2x^2), subtracted -1, and then divided each term by x^3. At the end I get the equation attached to this post. I don't know how to proceed...

I thought that f(9nthDerivative)(0)/9! is equal to the constants of the ninth term in the summation. i'm really stuck. Can someone help me?

Ps. sorry, I'm new to the forum and I don't know how to add equations :S
Note that the Maclaurin series formula is $\displaystyle f(x) = f(0) + f'(0) x + \frac{f''(0)}{2!} x^2 + \frac{f^{(3)}(0)}{3!} x^3 + .... + \frac{f^{(9)}(0)}{9!} x^9 + ....$

so get the coefficient of $\displaystyle x^9$ in the series you found and do with it what should be the obvious thing ....

3. Oooh, ok. I thought that I needed to find the ninth term! I got it now, it works! Thank you!