Power Rule

In calculus, the power rule is the following rule of differentiation.

Power Rule: For any real number $c$,

$$\frac{d}{dx} x^c = c x ^{c-1 }.$$

Using the rules of differentiation and the power rule, we can calculate the derivative of polynomials as follows:

Given a polynomial

$$f(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0,$$

the derivative of the polynomial is

$$\frac{d}{dx} f(x) = a_n \big(nx^{n-1}\big) + a_{n-1} (n-1) x^{n-2} + \cdots + a_1 (1) + a_0 (0).$$

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Proof: Using the addition and multiplication by a constant rules for differentiation, we have

$$\begin{aligned} \frac{d}{dx} f(x) &= \frac{d}{dx} \big(a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0\big)\\ & = \frac{d}{dx} \big(a_n x^n\big) + \frac{d}{dx} \big(a_{n-1}x^{n-1}\big) + \cdots + \frac{d}{dx} (a_1x) + \frac{d}{dx}(a_0) \\ &= a_n \frac{d}{dx} x^n + a_{n-1} \frac{d}{dx} x^{n-1} + \cdots + a_1 \frac{d}{dx} x + a_0 \frac{d}{dx}(1)\\ &= a_n \big(nx^{n-1}\big) + a_{n-1} (n-1) x^{n-2} + \cdots + a_1 (1) + a_0 (0), \end{aligned}$$

where the last line follows from the power rule. This proves the theorem. $_\square$

What is the derivative of $x^3?$

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Applying the power rule with $c = 3$, we have

$$\frac{ d}{dx} \big( x^3\big) = 3 x ^ 2 .\ _\square$$

What is $\frac{ d}{dx} \sqrt{x}?$

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Since $\sqrt{x} = x^{ \frac{1}{2} }$, we apply the power rule with $c = \frac{1}{2}$ to obtain

$$\frac{ d}{dx} \sqrt{x} = \frac{1}{2} x^{\frac{1}{2} - 1} = \frac{1}{2} x^{-\frac{1}{2}} = \frac{ 1 } { 2 \sqrt{x} }.\ _\square$$

What is $\frac{d}{dx} \left( \frac{1}{x^{5}} \right)?$

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We have

$$\frac{d}{dx} \left( \frac{1}{x^{5}} \right) = \frac{d}{dx} \left( x^{-5} \right) = -5x^{-6} = -\frac{5}{x^6}.\ _\square$$

Given the polynomial

$$f(x) = 2x^5 + 7x^3 + 4x^2 + x + 3,$$

what is $\frac{d}{dx} f(x)?$

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We use the power rule to calculate the derivative of polynomial $f(x):$

$$\begin{aligned} \frac{d}{dx} f(x) &= \frac{d}{dx} \big( 2x^5 + 7x^3 + 4x^2 + x + 3\big)\\ & = \frac{d}{dx} \big(2x^5\big) + \frac{d}{dx} \big(7x^3\big) + \frac{d}{dx} \big(4x^2\big) + \frac{d}{dx}(x) + \frac{d}{dx} (3) \\ &= 2 \frac{d}{dx} x^5 + 7 \frac{d}{dx} x^3 + 4 \frac{d}{dx} x^2 + \frac{d}{dx} x + 3 \frac{d}{dx}(1)\\ &= 2 \big(5x^4\big) + 7 \big(3 x^2\big) + 4 \big(2 x^1\big) + (1) + 3 (0)\\ &= 10x^4 + 21x^2 + 8x + 1.\ _\square \end{aligned}$$

Proof of Power Rule 1:

Using the identity $x^c = e^{c \ln x},$ we differentiate both sides using derivatives of exponential functions and the chain rule to obtain

$$\begin{aligned} \frac{d}{dx} x^c &= \frac{d}{dx} e^{c \ln x}\\ &= e^{c \ln x} \frac{d}{dx} (c \ln x) \\ &= e^{c \ln x} \left( \frac{c}{x} \right) \\ &= x^c \left( \frac{c}{x} \right) \\ &= cx^{c-1}.\ _\square \end{aligned}$$

Proof of Power Rule 2:

Recall the formal definition of a derivative

$$\frac{d}{dx} f(x) = \lim_{\Delta x \rightarrow 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}$$

and the binomial theorem $\displaystyle (a+b)^{n} = \sum_{k=0}^{n} {n \choose k}a^{k}b^{n-k}$.

Then when our $f(x) = x^{n}$,

$$\begin{aligned} \frac{d}{dx} x^n &= \lim_{\Delta x \rightarrow 0} \frac{(x+\Delta x)^{n}-x^{n}}{\Delta x}\\ &= \lim_{\Delta x \rightarrow 0} \frac{1}{\Delta x} \left( \sum_{k=0}^{n} {n \choose k}x^{k}\Delta x^{n-k}-x^{n} \right) \\ &= \lim_{\Delta x \rightarrow 0} \frac{1}{\Delta x}\sum_{k=1}^{n} {n \choose k}x^{k}\Delta x^{n-k} \\ &= \lim_{\Delta x \rightarrow 0} \sum_{k=1}^{n} {n \choose k}x^{k}\Delta x^{n-k-1}. \end{aligned}$$

At this point, the terms with $\Delta x$ disappear, and we are left with

$$\begin{aligned} \frac{d}{dx} x^n &= {n \choose n-1}x^{n-1} \\ &= nx^{n-1}.\ _\square \end{aligned}$$