The Power Rule for Derivatives

· Mohammad-Ali Bandzar

The power rule is used to find the derivative of polynomial terms. It tells us that the derivative of xnx^n will be nxn1nx^{n-1} for any real n.

Visualization

The easiest way to visualize this is to use it to take the derivative of x2x^2. We can let n=2 and use the rule we learned above to find the derivative to be 2x2x.

You can verify your solution by randomly drawing tangents on x2x^2 and seeing how the slope of those tangents will always be equal to 2x2x. Take x=0 as an example: the slope of our tangent at zero would be zero (a horizontal line) and 2(0)=0.

This is shown on our graph below where the red line is our original function, the blue line is our derivative, and the green line is our tangent at zero.

Power Rule Graph

Notation

The derivative is commonly indicated with either a “prime”/“apostrophe” (’) or with Leibniz notation we can write: ddx\frac{d}{dx}, which means the same thing.

So if f(x)=xnf(x) = x^n is our original, we could write the derivative as f(x)=nxn1f'(x) = nx^{n-1} or as: ddxf(x)=nxn1\frac{d}{dx} f(x) = nx^{n-1}.

Derivative of x\sqrt{x}

We can rewrite x\sqrt{x} as x12x^{\frac{1}{2}}. We can then apply our power rule by letting n=12n = \frac{1}{2}:

ddxx12\frac{d}{dx} x^{\frac{1}{2}}

=12x121= \frac{1}{2} x^{\frac{1}{2}-1}

=12x12= \frac{1}{2} x^{-\frac{1}{2}}

=12×x12= \frac{1}{2 \times x^{\frac{1}{2}}}

We can now conclude that:

ddxx=12×x12=12x\frac{d}{dx} \sqrt{x} = \frac{1}{2 \times x^{\frac{1}{2}}} = \frac{1}{2\sqrt{x}}

Memorization Trick

All you need to remember to do is to multiply by the power, then subtract 1 from the power (reduce the power by 1).

A Table of Examples

These examples are completely random and have been chosen to help you memorize the power rule:

f(x)derivative
x3x^33x23x^2
99x099x^000
5x25x^210x10x
x1x^{-1}x2-x^{-2}
2x0.32x^{0.3}0.6x0.70.6x^{-0.7}

Pro Tips

  • Works with any real number, not just integers
  • Multiply n by the coefficient of x, then subtract 1 from the power
  • The derivative will always be an order 1 less than the original
  • The derivative at a local min or max will always be zero as the tangent will be horizontal

THANKS FOR READING