y=x^3sinx,what's the 6th derivative of y at x = pi/6? question 1： $y=x^3\sin x$, what's $y^{\left(6\right)}\left(0\right)$ ?
I have solved it use Taylor's Formula in the follow way.
step1:
$y=\displaystyle \sum _{n=0}^{\infty }\frac{y^{\left(n\right)}\left(x_0\right)}{n!}\left(x-x_0\right)^n$
and x_0 = 0 for the question1,  So $y=\displaystyle \sum _{n=0}^{\infty }\frac{y^{\left(n\right)}\left(0\right)}{n!}\left(x\right)^n$
step2:
$y=x^3\sin x=x^3\left(x-\frac{1}{3!}x^3+\ldots \right)=x^4-\frac{1}{6}x^6+\ldots$
[PS:  $\sin x=\displaystyle \sum _{n=0}^{\infty }\:\frac{\left(-1\right)^n}{\left(2n+1\right)!}x^{2n+1}$]
step3:
$\dfrac{y^{\left(6\right)}\left(0\right)}{6!}\:=\:-\dfrac{1}{6}. \quad $
So $y^{\left(6\right)}\left(0\right)=-120$
The above method is so useful. But for question2 I find that I can't use the method above.Is there a way to use Taylor's formula to solve this problem, such as substitution?  I know leibniz formula can solve it but I want to know if Taylor's formula can be made better in this case?
Question2: $y=x^3\sin x$, what's $y^{\left(6\right)}\left(\dfrac{\pi }{6}\right)$?

Equivalently, we want $6![y^6](y+\pi/6)^3(\sqrt{3}\sin y+\cos y)/2$, where $[y^k]f(y)$ is the $y^k$ coefficient in $f(y)$. So the result is$$\begin{align}&360[y^6]\left(y^3+\frac{\pi}{2}y^2+\frac{\pi^2}{12}y+\frac{\pi^3}{216}\right)\left(1+y\sqrt{3}-\frac12y^2-\frac{\sqrt{3}}{6}y^3+\frac{1}{24}y^4+\frac{\sqrt{3}}{120}y^5-\frac{1}{720}y^6\right)\\&=360\left(-\frac{\sqrt{3}}{6}+\frac{\pi}{48}+\frac{\pi^2\sqrt{3}}{1440}-\frac{\pi^3}{216\times720}\right)\\&=-60\sqrt{3}+\frac{15\pi}{2}+\frac{\pi^2\sqrt{3}}{4}-\frac{\pi^3}{432}.\end{align}$$
