Skip to main content
Mathematics LibreTexts

3.2: Power Rule

  • Page ID
    88642
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    “I’ve got the power!” –Snap (German band)

    If you apply the definition of the derivative to several functions, you’ll see:

    \(\begin{array}{cc} \frac{d}{dx} x^0 & 0 \\ \frac{d}{dx} x^1 & 1\\ \frac{d}{dx} x^2 & 2x \\ \frac{d}{dx} x^3 & 3x^2 \\ \frac{d}{dx} x^4 & 4x^3 \end{array}\)

    See the pattern?

    \(\boxed{\frac{d}{dx} x^n = n x^{n-1}}\)

    As we’ll see in the next section, this even works for non-integers. The key is to multiply by the exponent, then decrease the exponent by one.

    The next rules say that constant multiples and addition work nicely.

    \(\boxed{\frac{d}{dx} c \cdot f(x) = c f'(x)}\)

    \(\boxed{\frac{d}{dx} [f(x) + g(x)] = f'(x) + g'(x)}\)

    The key is you can just worry about the derivative of each piece of a sum separately. Constant multiples “come along for the ride”. With only these three rules, you can now take the derivative of any polynomial. Check it out.

    Power

    Find the following derivatives.

    1. \(\frac{d}{dx} 3x^2\)

      To do this one, we use the power rule on the \(x^2\) part, and get \(2x\). However, we are also multiplying by \(3\), so the answer is multiplied by \(3\) as well. Hence the answer is \(3(2x) = \boxed{6x}\).

    2. \(\frac{d}{dx} x^3 + x\)

      By the power rule, we find \(\frac{d}{dx} x^3 = 3x^2\), and \(\frac{d}{dx} x\) is \(\frac{d}{dx} x^1\) which becomes \(1x^0\) by the power rule, which is \(1\). By the addition rule, we have \(\frac{d}{dx} x^3 + x = \boxed{3x^2 + 1}\).

    3. \(\frac{d}{dx} 2x^3 + 5\)

      You take the derivative of \(x^3\) and you have \(3x^2\). Times by \(2\), that leaves \(6x^2\). Okay, about the five? It is tempting to leave the five put, but actually \(\frac{d}{dx} 5 = 0\). Why? Well, it’s a constant, so it does not affect the slope. Hence we get \(\boxed{6x^2}\).

    More power rule examples

    Note that the power rule works with fractional and negative exponents as well! Here are some examples.

    Fractional

    Find quicklatex.com-9c6e8c37765e11f7e7d2af1e02b34506_l3.png.

    To apply the power rule in this case, we need to first multiply by the exponent (quicklatex.com-01a25cdafea4857da13bef49c076ac71_l3.png), then subtract one from the exponent quicklatex.com-bd358944f0b3f4a89caff1f8876f40e1_l3.png. Then we have

    quicklatex.com-dda1257ecda0a0470154dfbede5ef079_l3.png

    Negative

    Find \(\frac{d}{dx} 7x^{-2}\).

    In this problem, we just worry about the \(x^{-2}\) to start with. We multiply by the exponent \(({\color{red} -2})\), and then subtract one from that to get \({\color{blue} -2 - 1 = -3}\). Then we have

    \(\frac{d}{dx} 7x^{-2} = 7 ({\color{red} -2} x^

    ParseError: invalid DekiScript (click for details)
    Callstack:
        at (Bookshelves/Calculus/Informal_Calculus_with_Applications_to_Biological_and_Environmental_Sciences_(Seacrest)/03:_Rules_for_Derivatives/3.02:_Power_Rule), /content/body/div[2]/div[2]/div/p[3]/span, line 1, column 1
    
    ) = \boxed{-14 x^{-3}}\)

    Sometimes there are hidden fractional or negative exponents. Don’t let them fool you, they are just like the examples above. Just remember these rules:

    \(\boxed{\frac{1}{x^n} = x^{-n}}\) quicklatex.com-d59282ca237056fa481f3c95085e8557_l3.png

    Let’s see some examples.

    Roots

    Find

    1. \(\frac{d}{dx} \sqrt{x}\)

      This problem is much easier if we can rewrite the \(\sqrt{x}\). This is the same thing as quicklatex.com-7fd6591055d9e779ecf6ceb85ee1b282_l3.png, and hence we have

      quicklatex.com-7fe0d0da5dc85a80958c8534245c632b_l3.png

    2. \(\frac{d}{dx} 5\sqrt[3]{x}\)

      To solve this, it really helps to rewrite \(\sqrt[3]{x}\) as quicklatex.com-1ae6971de4610cd793d233336aa5295f_l3.png. Once you do that, this prolem is just the power rule and constant multiple rule:

      quicklatex.com-cb2d5632b97493e0b696e5c26d55fc1e_l3.png

    3. \(\frac{d}{dx} \sqrt[5]{x^3} + x^2 + 7\)

      Focus on the easy parts first: we know \(\frac{d}{dx} x^2 = 2x\), and we know that \(\frac{d}{dx} 7 = 0\). So we just need to figure out the \(\frac{d}{dx} \sqrt[5]{x^3}\). What is this? Well, we can rewrite this as quicklatex.com-9b61f2e99b17a6d69f6f57ff31fb4bdf_l3.png. So now it is just the power rule, and we multiply by quicklatex.com-ece565758c8366d16bfcbad1f651f424_l3.png and subtract to get quicklatex.com-173c25538bb19f97ea284d3154886f78_l3.png. Hence quicklatex.com-faa378ea6c49c4cd5654c9299e50042e_l3.png. Putting it all together:

    Powers of \(x\) in the denominator

    1. \(\frac{d}{dx} \frac{1}{x}\)

      We can rewrite \(\frac{d}{dx} \frac{1}{x} = \frac{d}{dx} \frac{1}{x^1}\) as \(\frac{d}{dx} x^{-1}\). Now we apply the power rule:

      \(\frac{d}{dx} \frac{1}{x} = \frac{d}{dx} x^{-1} = \boxed{-1 x^{-2}}.\)

    2. \(\frac{d}{dx} \frac{4}{x^2}\)

      We can think of \(4\) as just a constant out front, and hence we want \(\frac{d}{dx} 4 \left( \frac{1}{x^2} \right )\). We can then rewrite \(\frac{1}{x^2}\) as \(x^{-2}\). And we want \(\frac{d}{dx} 4(x^{-2})\). Using the power rule, we multiply by \(-2\) and subtract one, and we have

      \(\frac{d}{dx} \frac{4}{x^2} = \frac{d}{dx} 4x^{-2} = \boxed{-8 x^{-3}}.\)

    3. \(\frac{d}{dx} \frac{1}{\sqrt{x}}\)

      This combines the fractional and denominator stuff. We first rewrite \(\sqrt{x}\) as quicklatex.com-7fd6591055d9e779ecf6ceb85ee1b282_l3.png:

      quicklatex.com-06ccb62713936eb6a87a01b05bbd5d01_l3.png

      We then rewrite as a negative fractional exponent.

      quicklatex.com-344193b633d0447839d5a90b0911b5dc_l3.png

      Finally, we use the power rule.

      quicklatex.com-d3feeec8da7bee4a9a11d4088070c428_l3.png


    This page titled 3.2: Power Rule is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Tyler Seacrest via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.