100 derivado de la $(1-2x)^{2/3}$ a punto de $x=0$

Question

$$\frac{\mathrm d^{100}}{\mathrm dx^{100}} (1-2x)^{2/3}$$ Sin Taylor.

Yo relé no tienen ninguna idea de cómo Generales de uso regla de Leibniz en este caso.

Answer 1

$$\frac d{dx} f(x)^n=nf(x)^{n-1}f'(x)$$

Si $f'(x)=k$ para algunas constantes $k$

$$\frac{d^m}{dx^m}f(x)^n=n(n-1)...(n-m+1)f(x)^{n-m} k^m$$

Answer 2

Aviso:

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

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Así: $$\frac{\text{d}^{100}}{\text{d}x^{100}}\left[\left(1-2x\right)^{\frac{2}{3}}\right]=(-2)^{100}\cdot(1-2x)^{\frac{2}{3}-100}\cdot\frac{\Gamma\left(\frac{5}{3}\right)}{\Gamma\left(\frac{5}{3}-100\right)}=$$ $$2^{100}\cdot(1-2x)^{-\frac{298}{3}}\cdot\frac{\Gamma\left(\frac{5}{3}\right)}{\Gamma\left(-\frac{295}{3}\right)}$$

Y en el punto de $x=0$:

$$2^{100}\cdot(1-2\cdot0)^{-\frac{298}{3}}\cdot\frac{\Gamma\left(\frac{5}{3}\right)}{\Gamma\left(-\frac{295}{3}\right)}=2^{100}\cdot(1-0)^{-\frac{298}{3}}\cdot\frac{\Gamma\left(\frac{5}{3}\right)}{\Gamma\left(-\frac{295}{3}\right)}=$$ $$2^{100}\cdot(1)^{-\frac{298}{3}}\cdot\frac{\Gamma\left(\frac{5}{3}\right)}{\Gamma\left(-\frac{295}{3}\right)}=2^{100}\cdot1\cdot\frac{\Gamma\left(\frac{5}{3}\right)}{\Gamma\left(-\frac{295}{3}\right)}=\frac{2^{100}\Gamma\left(\frac{5}{3}\right)}{\Gamma\left(-\frac{295}{3}\right)}\approx-1.37416\times10^{184}$$