# The Derivative Power Rule $\frac d {dx}\bigg[x^n\bigg] = nx^{n-1}$ ^1osjwp <div> <img src="https://d18l82el6cdm1i.cloudfront.net/uploads/PWXHIhOvhj-c3q1p4.svg?width=800" style="background: white; padding: 10px; width: 250px; margin:auto; display:block" /> </div> ## Proof Using Limits $ \begin{align} \frac d {dx} x^n &= \lim_{x \to a} \frac {x^n - a^n}{x-a} \\ &= \lim_{x \to a} \frac {(x-a)\sum_{k=0}^{n-1} x^{n-k-1}a^k}{x-a} && \text{see note 1} \\ &= \lim_{x \to a} \sum_{k=0}^{n-1} x^{n-k-1}a^k \\ &= \sum_{k=0}^{n-1} a^{n-k-1}a^k && \text{see note 2} \\ &= \sum_{k=0}^{n-1} a^{n-1} \\ \frac d {dx} x^n &= na^{n-1} \end{align} $ Notes: 1. As per [[The General Formula for the Difference of Two Nth Powers]] 2. See [[The Limit Power Rule]] ## Proof Using The Binomial Theorem Per [[Binomial Theorem]]: ![[Binomial Theorem#^knv0jb]] so $(x+dx)^n = x^n + nx^{n-1}dx + \sum_{k=2}^n \binom n k x^{n-k}dx^k$ From here it follows that $dy$ is everything on the right side except for the first term: $dy=nx^{n-1}dx + \sum_{k=2}^n \binom n k x^{n-k}dx^k$ When we divide both sides by $dx$ we get: $\frac {dy}{dx} = nx^{n-1} + \sum_{k=2}^n \binom n k x^{n-k}dx^{k-1}$ The second part of this expression can be ignored as $dx$ tends to 0. ## Deriving the Derivative of a Square Root The power rule can also be used to determine the derivative of a square root $\frac d {dx} \bigg[ \sqrt x \bigg] = \frac 1 {2\sqrt x}$ ^gv5xph $\frac d {dx} \bigg[ x^{\frac 1 2} \bigg] = \frac 1 2 x^{-\frac 1 2} = \frac 1 {2\sqrt x}$ You can reason about this geometrically by thinking of a square whose area is $x$ and side $\sqrt x$. The change of the area of the square $dx$, for an increase of its side $d\sqrt x$, is $2\sqrt x\, d\sqrt x + (d\sqrt x)^2$. This corresponds to the area of the two additional rectangles and the additional corner square. (important to note here that $x$ is the area!) $dx = 2\sqrt x \, d\sqrt x + (d\sqrt x)^2$ We can divide both sides by $d\sqrt x$: $ \frac {dx} {d\sqrt x} = 2\sqrt x + d \sqrt x$ Because $d\sqrt x$ tends to 0 we can ignore the second part of the right side. This gives us the change of $x$ for a change of $\sqrt x$, but we're actually interested in the reciprocal! We can then flip the ratio to come up with the desired derivative: $\frac {d \sqrt x} {dx} = \frac 1 {2\sqrt x }$