Calculate $f^{(25)}(0)$ for $f(x)=x^2 \sin(x)$ Calculate $f^{(25)}(0)$ for $f(x)=x^2 \sin(x)$.
The answer is too short for me to understand, and the answer is
$- 25 \cdot 24 \cdot 8^{23}$
 A: use the leibnitz rule to find the  nth derivative of the function.
$$\frac{d^n}{dx^n}(uv) = ^nC_0u \frac{d^n}{dx^n} (v)+^nC_1 \frac{d}{dx} (u) \frac{d^{n-1}}{dx^{n-1}} v+...... ^nC_n \frac{d^n}{dx^n} (u)v$$
here $  u=x^2$ and $v=\sin x$
also note
$$ \frac{d^n}{dx^n} (\sin x)=sin(x+n\frac{\pi}{2})$$
on solving you get
$$f^{n}(x)=x^2\sin(x+n\frac{\pi}{2})+2nx\sin(x+(n-1)\frac{\pi}{2})+n(n-1)\sin (x+(n-2)\frac{\pi}{2})$$
A: The answer you have stated is wrong. It should be $-25*24$. Note that in the Taylor expansion for $f(x)$ the only non-zero term that matters is $-x^{25}/23!$.  Differentiate this last expression $25$ terms and plug in zero, you will get 
$$f^{(25)}(0) = -25!/23! = -600.$$
A: $$f(x) = x^2\sin x$$
$$ = x^2 \left(x - \frac{x^3}{3!} +...- \frac{x^{23}}{23!} +...\right)$$
$$ = -\frac{x^{25}}{23!} +...$$
$$f^{(25)}(x) = -\frac{25!}{23!} + ...$$
$$f^{(25)}(0) = -600$$
WolframAlpha verifies that the $8^{23}$ should not be there.
A: Here is an approach. Follow the steps
i) use the identity $\sin x = \frac{e^{ix}-e^{-ix} }{2i} $
ii) use the product rule for differentition.
Note that the function $(x^2)^{(m) }$ vanishes for $m=3$.  Chech this related problem.
