I am able to find the sixth derivative of $\cos(x^2)$ by simply replacing the $x$ in the Taylor series for $\cos(x)$ with $x^2$ but beyond simple substitutions, I am struggling...
Thanks for any help!
$\endgroup$ 43 Answers
$\begingroup$$\displaystyle f(x)=\frac{1}{2}+\frac{1}{2}\cos 2x$.
$\displaystyle f^{(6)}(x)=\frac{1}{2}(2)^6(-\cos 2x)=-32\cos2x$.
$\endgroup$ $\begingroup$Since you need to differentiate $6$ times, you can just brute force it. Indeed $$ \begin{align} f(x)&=\cos^2x\\ f'(x)&=-2\cos x\sin x=-\sin(2x)\\ f''(x)&=-2\cos(2x)\\ &\vdots \end{align} $$ and so on. The rest is just remembering the trig derivatives and using the chain rule.
$\endgroup$ $\begingroup$Hint: Linearise first: $$\cos^2x=\frac12(1+\cos 2x),$$ and differentiate $\cos 2x$ six times, remembering that $$(\cos u)'=\cos\Bigl(u+\frac\pi2\Bigr)u'=\cos\Bigl(2x+\frac\pi2\Bigr)\cdot 2.$$
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