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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:

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


Natural log method

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

Let $ y=x^r $ . Then:

$ \begin{align} &\ln(y)=\ln(x^r)=r\ln(x)\\ &\frac{d}{dx}\big(\ln(y)\big)=\frac{d}{dx}\big(r\ln(x)\big)\\ &\frac{y'}{y}=\frac{r}{x}\\ &y'=\frac{ry}{x}=\frac{rx^r}{x}=rx^{r-1} \end{align} $
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