Trick for calculating the 5th order derivative of $(2x^3+1)e^{x^2}$ evaluated at $0$

Question

Are there any tricks for calculating the 5th order derivative of $(2x^3+1)e^{x^2}$ evaluated at $0$?
I guess it involves binomial expansion but it still seems too complicated.

Answer 1

using serie expansion $$let : \;f(x)=(1+2x^3)e^{x^2}$$ $$ e^{x^2}＝1＋x^2＋\frac {x^4}{2!}＋{\rm O}(x^5)\\~\\f(x)=(1＋2x^3)e^{x^2}＝(1＋2x^3)(1＋x^2＋\frac {x^4}{2!})＋＋{\rm O}(x^5)＝1＋x^2+2x^3+\frac {1}{2}x^4＋2x^5+{\rm O}(x^5)$$

And therfore : $$f(x)=f(0)＋\frac {f^{(1)}(0)}{1!}x＋\frac {f^{(2)}(0)}{2!}x^2＋\frac {f^{(3)}(0)}{3!}x^3＋\frac {f^{(4)}(0)}{4!}x^4＋\frac {f^{(5)}(0)}{5!}x^5＋{\rm O}(x^5)$$

We note that: $$\frac {f^{(5)}(0)}{5!}＝2\implies\;f^{(5)}(0)＝2×5!=240$$

So, the fifth-degree derivative at x=0 is:240

Answer 2

One way to do it think about $\frac{d^5}{dx^5}(fg)$ if you derive once you get $f'g+fg'$ twice you get $f''g+2f'g'+g''$ three times $f'''g+3f''g'+3f'g''+fg'''$ in general

$$\frac{d^n}{dx^n}(fg) = \sum_{k=0}^n \binom{n}{k} f^{(k)}g^{(n-k)}$$ Follows the binomial expansion. The good thing about your case is $(2x^3+1)$ vanish after 4 derivatives. Also, after the first derivative when you plug $0$ it will be zero! so what are the remaining terms? On the other hand after the first derivative of $e^{x^2}$ you get zero when you plug $x=0$ so?