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 $y^{\left(6\right)}\left(\dfrac{\pi }{6}\right)$?
 A: Thanks for @InterstellarProbe's helping me solve this problem.
$y=x^3sinx$
step1:
Let : $x=u+\frac{\pi }{6}$
then
$y=\sum _{n=0}^{\infty }\:\frac{y^{\left(n\right)}\left(0\right)}{n!}u^n$
and
$y=\left(u+\frac{\pi }{6}\right)^3sin\left(u+\frac{\pi }{6}\right)
 =\left(u^3+\frac{\pi }{2}u^2+\frac{\pi^2 }{12}u+\frac{\pi \:^3}{6^3}\right)\left(sin\left(u\right)cos\left(\frac{\pi }{6}\right)+cos\left(u\right)sin\left(\frac{\pi }{6}\right)\right)
=\left(u^3+\frac{\pi }{2}u^2+\frac{\pi^2 }{12}u+\frac{\pi ^3}{6^3}\right)\left(\frac{\sqrt{3}}{2}\left(u-\frac{1}{3!}u^3+\frac{1}{5!}u^5-\frac{1}{7!}u^7+...\right)+\frac{1}{2}\left(1-\frac{1}{2!}u^2+\frac{1}{4!}u^4-\frac{1}{6!}u^6+...\right)\right)\:$
step2:
$\frac{y^{\left(6\right)}\left(0\right)}{6!}u^6=u^3\cdot \left(\frac{\sqrt{3}}{2}\cdot \left(-\frac{1}{6}\right)u^3\right)+\frac{\pi }{6}u^2\cdot \left(\frac{1}{2}\cdot \frac{1}{4!}u^4\right)+\frac{\pi^2 }{12}u\cdot \left(\frac{\sqrt{3}}{2}\cdot \frac{1}{5!}u^5\right)+\frac{\pi ^3}{6^3}\cdot \left((\frac{1}{2})(-\frac{1}{6!})u^6\right)$
So
$\frac{y^{\left(6\right)}\left(0\right)}{6!}=-\frac{\sqrt{3}}{12}+\frac{\pi }{4\cdot 4!}+\frac{\sqrt{3}\pi }{24\cdot 5!}-\frac{\pi ^3}{2*6!\cdot 6^3}$
and the anwser is 
$y^{\left(6\right)}\left(0\right)=\left(-\frac{\sqrt{3}}{12}+\frac{\pi \:}{4\cdot \:4!}+\frac{\sqrt{3}\pi^2 \:}{24\cdot \:5!}-\frac{\pi \:^3}{2*6!\cdot \:6^3}\right)\cdot 6!$
A: 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$$\begin{align}&360[y^6]\left(y^3+\frac{\pi}{2}y^2+\frac{\pi^2}{12}y+\frac{\pi^3}{216}\right)\left(1+y\sqrt{3}-\frac12y^2-\frac{\sqrt{3}}{6}y^3+\frac{1}{24}y^4+\frac{\sqrt{3}}{120}y^5-\frac{1}{720}y^6\right)\\&=360\left(-\frac{\sqrt{3}}{6}+\frac{\pi}{48}+\frac{\pi^2\sqrt{3}}{1440}-\frac{\pi^3}{216\times720}\right)\\&=-60\sqrt{3}+\frac{15\pi}{2}+\frac{\pi^2\sqrt{3}}{4}-\frac{\pi^3}{432}.\end{align}$$
A: The $n-$th derivative of $f(x)g(x)$expands with the same binomial cefficints as $(a+b)^n$, so $$(f(x)g(x)^n)=f^{(n)}(x)g(x)+nf^{(n-1)}(x)g'(x)+...+f(x)g^{(n)}(x).$$ Thus $$(x^3\sin(x))^{vi}={6 \choose 3}(x^3)'''\sin'''(x)$$
$$+{6 \choose 4}(x^3)''\sin''''(x)$$ $$+{6 \choose 5}(x^3)'\sin'''''(x)$$ $$+{6 \choose 6}(x^3)\sin''''''(x)$$ $$= 20(6)(-\cos(x)$$
$$+15(6x)\sin(x)$$ $$+6(3x^2)\cos(x)$$ $$+(x^3)(-\sin(x))$$ Now put $x=\frac{\pi}{6}.$
