Expand the function $(\cos(x^3))^{-1/2}$ into the Taylor series around $x = 0$ up to $o(x^6)$ Expand the function $f(x) = (\cos(x^3))^{-1/2}$ into the Taylor series around $x = 0$ up to $o(x^6)$
I started with a standard approach finding derivatives of this function.
$f(0)=1$, $f'(0)=0$. After that it becomes messy, but I feel like all other derivatives should be $0$ too. Is it the right approach?
 A: Hint (for an easier way). Since $\cos(t)=1-t^2/2+o(t^2)$ it follows that
$$(\cos(x^3))^{-1/2}=\left(1-\frac{(x^3)^2}{2}+o(x^6)\right)^{-1/2}.$$
Now recall the expansion $(1+t)^a=1+at +o(t)$ at $t=0$. Can you take it from here?
A: Since the Taylor series of $\cos\left(x^3\right)$ around $0$ is $1-\frac{x^6}2+0(x^6)$, the Taylor series of $\sqrt{\cos\left(x^3\right)}$ around $0$ is $1-\frac{x^6}4+o(x^6)$ and therefore the Taylor series of $\frac1{\sqrt{\cos\left(x^3\right)}}$ around $0$ is $1+\frac{x^6}4+o(x^6)$.
A: A less standard approach:
As the cosine is an even function, it is lacking odd powers in its development, and the second term of $\cos x^3$ is in $x^6$. Then composing with $(1+x)^{-1/2}\approx 1-\dfrac x2$, no term of lower degree can appear.
To obtain the coefficient, we could differentiate $\cos^{-1/2}t$ twice and use the quadratic coefficient. But it is more efficient to differentiate $\cos^{-1/2}\sqrt t$ once.
We have
$$(\cos^{-1/2}\sqrt t)'=-\frac12\frac{-\sin\sqrt t}{2\sqrt t}\cos^{-3/2}\sqrt t$$ so that
$$(\cos t^3)^{-1/2}=1+\frac{x^6}4+o(x^6).$$
