“I’ve got the power!” –Snap (German band)
If you apply the definition of the derivative to several functions, you’ll see:
[latex]\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}[/latex]
See the pattern?
[latex]\boxed{\frac{d}{dx} x^n = n x^{n-1}}[/latex]
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.
[latex]\boxed{\frac{d}{dx} c \cdot f(x) = c f'(x)}[/latex]
[latex]\boxed{\frac{d}{dx} [f(x) + g(x)] = f'(x) + g'(x)}[/latex]
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.
Find the following derivatives.
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[latex]\frac{d}{dx} 3x^2[/latex]
To do this one, we use the power rule on the [latex]x^2[/latex] part, and get [latex]2x[/latex]. However, we are also multiplying by [latex]3[/latex], so the answer is multiplied by [latex]3[/latex] as well. Hence the answer is [latex]3(2x) = \boxed{6x}[/latex].
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[latex]\frac{d}{dx} x^3 + x[/latex]
By the power rule, we find [latex]\frac{d}{dx} x^3 = 3x^2[/latex], and [latex]\frac{d}{dx} x[/latex] is [latex]\frac{d}{dx} x^1[/latex] which becomes [latex]1x^0[/latex] by the power rule, which is [latex]1[/latex]. By the addition rule, we have [latex]\frac{d}{dx} x^3 + x = \boxed{3x^2 + 1}[/latex].
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[latex]\frac{d}{dx} 2x^3 + 5[/latex]
You take the derivative of [latex]x^3[/latex] and you have [latex]3x^2[/latex]. Times by [latex]2[/latex], that leaves [latex]6x^2[/latex]. Okay, about the five? It is tempting to leave the five put, but actually [latex]\frac{d}{dx} 5 = 0[/latex]. Why? Well, it’s a constant, so it does not affect the slope. Hence we get [latex]\boxed{6x^2}[/latex].
More power rule examples
Note that the power rule works with fractional and negative exponents as well! Here are some examples.
To apply the power rule in this case, we need to first multiply by the exponent ([latex]{\color{red} 3/2}[/latex]), then subtract one from the exponent [latex]{\color{blue} 3/2 - 1 = 1/2}[/latex]. Then we have
[latex]\frac{d}{dx} x^{3/2} = {\color{red} \frac{3}{2}} x^{{\color{blue} 1/2}} = \boxed{\frac{3}{2} x^{1/2}}[/latex]
In this problem, we just worry about the [latex]x^{-2}[/latex] to start with. We multiply by the exponent [latex]({\color{red} -2})[/latex], and then subtract one from that to get [latex]{\color{blue} -2 - 1 = -3}[/latex]. Then we have
[latex]\frac{d}{dx} 7x^{-2} = 7 ({\color{red} -2} x^{{\color{blue} -3}}) = \boxed{-14 x^{-3}}[/latex]
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:
[latex]\boxed{\frac{1}{x^n} = x^{-n}}[/latex] [latex]\boxed{\sqrt[m]{x^n} = \sqrt[m]{x}^n = x^{n/m}}[/latex]
Let’s see some examples.
Find
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[latex]\frac{d}{dx} \sqrt{x}[/latex]
This problem is much easier if we can rewrite the [latex]\sqrt{x}[/latex]. This is the same thing as [latex]x^{1/2}[/latex], and hence we have
[latex]\frac{d}{dx} \sqrt{x} = \frac{d}{dx} x^{1/2} = \frac{1}{2} x^{1/2 - 1} = \boxed{\frac{1}{2} x^{-1/2}}.[/latex]
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[latex]\frac{d}{dx} 5\sqrt[3]{x}[/latex]
To solve this, it really helps to rewrite [latex]\sqrt[3]{x}[/latex] as [latex]x^{1/3}[/latex]. Once you do that, this prolem is just the power rule and constant multiple rule:
[latex]\frac{d}{dx} 5 \sqrt[3]{x} = 5 x^{1/3} = 5 \left( \frac13 \right) x^{1/3 - 1} = \boxed{\frac53 x^{-2/3}}.[/latex]
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[latex]\frac{d}{dx} \sqrt[5]{x^3} + x^2 + 7[/latex]
Focus on the easy parts first: we know [latex]\frac{d}{dx} x^2 = 2x[/latex], and we know that [latex]\frac{d}{dx} 7 = 0[/latex]. So we just need to figure out the [latex]\frac{d}{dx} \sqrt[5]{x^3}[/latex]. What is this? Well, we can rewrite this as [latex]\frac{d}{dx} x^{3/5}[/latex]. So now it is just the power rule, and we multiply by [latex]3/5[/latex] and subtract to get [latex]-2/5[/latex]. Hence [latex]\frac{d}{dx} x^{3/5} = \frac{3}{5} x^{-2/5}[/latex]. Putting it all together:
[latex]\frac{d}{dx} \sqrt[5]{x^3} + x^2 + 7 = \boxed{\frac{3}{5} x^{-2/5} + 2x}.[/latex]
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[latex]\frac{d}{dx} \frac{1}{x}[/latex]
We can rewrite [latex]\frac{d}{dx} \frac{1}{x} = \frac{d}{dx} \frac{1}{x^1}[/latex] as [latex]\frac{d}{dx} x^{-1}[/latex]. Now we apply the power rule:
[latex]\frac{d}{dx} \frac{1}{x} = \frac{d}{dx} x^{-1} = \boxed{-1 x^{-2}}.[/latex]
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[latex]\frac{d}{dx} \frac{4}{x^2}[/latex]
We can think of [latex]4[/latex] as just a constant out front, and hence we want [latex]\frac{d}{dx} 4 \left( \frac{1}{x^2} \right )[/latex]. We can then rewrite [latex]\frac{1}{x^2}[/latex] as [latex]x^{-2}[/latex]. And we want [latex]\frac{d}{dx} 4(x^{-2})[/latex]. Using the power rule, we multiply by [latex]-2[/latex] and subtract one, and we have
[latex]\frac{d}{dx} \frac{4}{x^2} = \frac{d}{dx} 4x^{-2} = \boxed{-8 x^{-3}}.[/latex]
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[latex]\frac{d}{dx} \frac{1}{\sqrt{x}}[/latex]
This combines the fractional and denominator stuff. We first rewrite [latex]\sqrt{x}[/latex] as [latex]x^{1/2}[/latex]:
[latex]\frac{d}{dx} \frac{1}{\sqrt{x}} = \frac{d}{dx} \frac{1}{x^{1/2}}.[/latex]
We then rewrite as a negative fractional exponent.
[latex]\frac{d}{dx} \frac{1}{\sqrt{x}} = \frac{d}{dx} x^{-1/2}[/latex]
Finally, we use the power rule.
[latex]\frac{d}{dx} \frac{1}{\sqrt{x}} = \boxed{-\frac{1}{2} x^{-3/2}}.[/latex]