# Compute the 10th derivative of ...

#### s3a

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!

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#### Random Variable

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

Last edited:
• s3a

#### 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)

#### lilaziz1

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 $$\displaystyle \frac{6^{6} 10!}{6!}$$

• s3a

#### Random Variable

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

• s3a

#### s3a

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