100th derivative of $(1-2x)^{2/3}$ at point $x=0$ $$\frac{\mathrm d^{100}}{\mathrm dx^{100}} (1-2x)^{2/3}$$
Without Taylor.
I relay don't have any idea how to use General Leibniz rule in this case.
 A: $$\frac d{dx} f(x)^n=nf(x)^{n-1}f'(x)$$
If $f'(x)=k$ for some constant $k$
$$\frac{d^m}{dx^m}f(x)^n=n(n-1)...(n-m+1)f(x)^{n-m} k^m$$
A: Notice:


*

*$$\frac{\partial^n}{\partial x^n}\left[\left(1-2x\right)^{\frac{2}{3}}\right]=(-2)^n\cdot(1-2x)^{\frac{2}{3}-n}\cdot\frac{\Gamma\left(\frac{5}{3}\right)}{\Gamma\left(\frac{5}{3}-n\right)}$$



So:
$$\frac{\text{d}^{100}}{\text{d}x^{100}}\left[\left(1-2x\right)^{\frac{2}{3}}\right]=(-2)^{100}\cdot(1-2x)^{\frac{2}{3}-100}\cdot\frac{\Gamma\left(\frac{5}{3}\right)}{\Gamma\left(\frac{5}{3}-100\right)}=$$
$$2^{100}\cdot(1-2x)^{-\frac{298}{3}}\cdot\frac{\Gamma\left(\frac{5}{3}\right)}{\Gamma\left(-\frac{295}{3}\right)}$$
And in the point $x=0$:
$$2^{100}\cdot(1-2\cdot0)^{-\frac{298}{3}}\cdot\frac{\Gamma\left(\frac{5}{3}\right)}{\Gamma\left(-\frac{295}{3}\right)}=2^{100}\cdot(1-0)^{-\frac{298}{3}}\cdot\frac{\Gamma\left(\frac{5}{3}\right)}{\Gamma\left(-\frac{295}{3}\right)}=$$
$$2^{100}\cdot(1)^{-\frac{298}{3}}\cdot\frac{\Gamma\left(\frac{5}{3}\right)}{\Gamma\left(-\frac{295}{3}\right)}=2^{100}\cdot1\cdot\frac{\Gamma\left(\frac{5}{3}\right)}{\Gamma\left(-\frac{295}{3}\right)}=\frac{2^{100}\Gamma\left(\frac{5}{3}\right)}{\Gamma\left(-\frac{295}{3}\right)}\approx-1.37416\times10^{184}$$
