power rule

The formula will be proved using binomial theorem.

Let \[y=x^{r}\], where \[r\in\mathbb{N}_{0}\]. Then

\begin{align*} y^{\prime}&=\lim_{h\to0}\frac{(x+h)^{r}-x^{r}}{h}\\ &=\lim_{h\to0}\frac{1}{h}\left[ \left(x^{r}+\binom{r}{1}x^{r-1}h+\binom{r}{2}x^{r-2}h^{2}+\cdots+\binom{r}{r}h^{r}\right)-x^{r} \right]\\ &=\lim_{h\to0}\left[ \binom{r}{1}x^{r-1}+\binom{r}{2}x^{r-2}h+\cdots+\binom{n}{n}h^{r-1} \right]\\ &=rx^{r-1} \end{align*}

As the term \[x^{r}\] cancels out each other, and we multiply \[\frac{1}{h}\] into the binomial expansion.

Using the reciprocal rule, we let \[r=-m\], thus \[x^{r}=x^{-m}=\frac{1}{x^{m}}\].

\begin{align*} (x^{r})^{\prime}&=\left( \frac{1}{x^{m}} \right)^{\prime}\\ &=-\frac{(x^{m})^{\prime}}{(x^{m})^{2}}\\ &=-\frac{mx^{m-1}}{x^{2m}}\\ &=-mx^{-m-1}\\ &=nx^{n-1}\\ \end{align*}

Since \[x=e^{\ln x}\], we can say: \[x^{r}=\left( e^{\ln x} \right)^{r}=e^{r\ln x}\].

We take the derivative of both sides, \[\frac{d}{dx}x^{r}=\frac{d}{dx}e^{r\ln x}\]:

\begin{align*} \frac{d}{dx}x^{r}&=\frac{d}{dx}e^{r\ln x}\\ &=e^{r\ln x}\cdot \frac{d}{dx}(r\ln x)\\ &=e^{r\ln x}\left( \frac{r}{x} \right)\\ &=x^{r}\left( \frac{r}{x} \right)\\ &=rx^{r-1}\\ \end{align*}