Compute the 10th derivative of ...

**s3a** · May 17th 2010, 11:22 AM

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!
Thanks in advance!

**Random Variable** · May 17th 2010, 12:23 PM

Start with the known Maclaurin series of $\cos x$

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

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

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

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

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

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

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

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

**s3a** · May 17th 2010, 12:48 PM

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)

**lilaziz1** · May 17th 2010, 12:53 PM

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