Find 9th Derivative of f(x) = cos(6x^4)-1 at x=0 | Maclaurin Series Homework

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Homework Statement




(1 pt) Compute the 9th derivative of:
f(x) = \frac{cos(6x^4)-1}{x^7}
at x=0.



Homework Equations



Hint: Use the MacLaurin series for f(x).

The Attempt at a Solution



I have tried many weird ways and cannot come up with the correct numerical answer. I've gotten 0 each time and it still comes up as wrong. Please help point me in the correct direction, even if it is only a verbal explanation!

Thanks so much
 
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The only term in the expansion that would contribute to a 9th derivative would be the term containing x^9. If you look up the Maclaurin series of cos you'll it can be written as a sum of even powers of it's argument. The fourth power of 6x^4 contains an x^16. 16-7=9. Is that enough of a hint.
 
Yes, thank you very much. I think I can work it out now!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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