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\frac{d}{dx}(x^r)=rx^{r-1} , for any real number r .

Proofs

Binomial theorem method

Let f(x)=x^r . Then:

f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}
f'(x)=\lim_{h\to0}\frac{(x+h)^r-x^r}{h}
f'(x)=\lim_{h\to0}\frac{(x^r+rx^{r-1}h+\cdots+rxh^{r-1}+h^r)-x^r}{h}
f'(x)=\lim_{h\to0}\frac{rx^{r-1}h+\cdots+rxh^{r-1}+h^r}{h}
f'(x)=\lim_{h\to0}\frac{h(rx^{r-1}+\cdots+rxh^{r-2}+h^{r-1}}{h}
f'(x)=\lim_{h\to0}\Big[rx^{r-1}+\cdots+rxh^{r-2}+h^{r-1}\Big]
f'(x)=rx^{r-1}


Natural log method

Prerequisite: \frac{d}{dx}(\ln(u))=\frac{u'}{u}

Let y=x^r . Then:

\ln(y)=\ln(x^r)=r\ln(x)
\frac{d}{dx}(\ln(y))=\frac{d}{dx}(r\ln(x))
\frac{y'}{y}=\frac{r}{x}
y'=\frac{ry}{x}=\frac{rx^r}{x}=rx^{r-1}

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