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

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

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 \(f(x)\):

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

Proof of Power Rule: Using the identity \( x^c = e^{c \ln x},\) we differentiate both sides using derivatives of exponential functions and the chain rule to obtain

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

Proof of Power Rule: 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 \(f(x) = x^{n}\),

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

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

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