# Thread: Compute the 10th derivative of ...

1. ## Compute the 10th derivative of ...

The question is attached. I know the power series representation of cos(x) and therefore can get cos(6x^2)/x^2 but I don't know what to do about the -1.

Any help would be greatly appreciated!

2. Start with the known Maclaurin series of $\displaystyle \cos x$

$\displaystyle \cos x = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \frac{x^{6}}{6!} + ...$

then $\displaystyle \cos 6x^{2} = 1 - \frac{(6x^{2})^{2}}{2!} + \frac{(6x^{2})^{4}}{4!} - \frac{(6x^{2})^{6}}{6!} + ... = 1 - \frac{6^{2}x^{4}}{2!} + \frac{6^{4}x^{8}}{4!} - \frac{6^{6}x^{12}}{6!} + ...$

and $\displaystyle \cos 6x^{2} -1 = -\frac{6^{2}x^{4}}{2!} + \frac{6^{4}x^{8}}{4!} - \frac{6^{6}x^{12}}{6!} + ...$

finally $\displaystyle \frac{\cos 6x^{2}-1}{x^{2}} = -\frac{6^{2}x^{2}}{2!} + \frac{6^{4}x^{6}}{4!} - \frac{6^{6}x^{10}}{6!} + ...$

For a Macluarin series, the coefficent of the $\displaystyle x^{10}$ term is $\displaystyle \frac{f^{(10)}(0)}{10!}$

The coefficient of the $\displaystyle x^{10}$ term in the above Macluarin series is $\displaystyle -\frac{6^{6}}{6!}$

so $\displaystyle \frac{f^{(10)}(0)}{10!} = -\frac{6^{6}}{6!}$

EDIT: and $\displaystyle f^{(10)}(0) = - \frac{6^{6} 10!}{6!} = - 235146240$

3. How do you get from the before-last step to the last one? (From the one that says "so" to the one with the emoticon)

4. the 10! came from the exponent on $\displaystyle \frac{6^{6}x^{10}}{6!}$. If you were to take the derivative of that 10 times, you would get $\displaystyle \frac{6^{6} 10!}{6!}$

5. Originally Posted by s3a
How do you get from the before-last step to the last one? (From the one that says "so" to the one with the emoticon)
It should be $\displaystyle f^{(10)} (0) = - \frac{6^{6} 10!}{6!}$

6. Oh that 10! was also tricky but I get it now! Thanks!