Course:MATH110/Archive/2010-2011/003/Notes/Power Rule

The Power Rule

The power rule is usually the first formula that is proven. It tells you the derivative of one of the simplest type of function: monomials.

The Power Rule
Let $n$ be an integer, positive or negative or even a fraction, then The derivative of $x^n$ is $n{x}^{n-1}$.

It is quite easy to prove this formula when $n$ is a positive integer, for negative integers or fractions it is slightly more tricky. A proof will be shown below. But first, let's give some examples.

Examples

$f(x)$	$f^{\prime }(x)$
$x^3$	$3x^2$
$x^{143}$	$143x^{142}$
$x=x^1$	$1x^0=1$
$\frac{1}{x^2}=x^{-2}$	$-2x^{-3}=-\frac{2}{x^3}$
$\sqrt{x}=x^{1/2}$	$\frac{1}{2}{x}^{-1/2}=\frac{1}{2\sqrt{x}}$
$x^{3/2}=\sqrt{x^3}$	$\frac{3}{2}{x}^{1/2}=\frac{3}{2}\sqrt{x}$
$x^{7/5}=\sqrt[5]{x^7}$	$\frac{7}{5}{x}^{2/5}=\frac{7}{5}\sqrt[5]{x^2}$
$x^{-7/5}=\frac{1}{\sqrt[5]{x^7}}$	$-\frac{7}{5}{x}^{-12/5}=-\frac{7}{5\sqrt[5]{x^{12}}}$

A proof of the Power Rule

Here's a proof for the power rule when the exponent is a positive integer. Instead of relying on heavy algebra, this proof relies on a more abstract concept: the proof by induction. What we'll do here is to prove the power rule for $n=1$ and then prove that if the formula is true for some $n=k$ then it is true for the next power $n=k+1$. So this works like a computer program where we show that the first step is true and then prove that we can always go from there to the next step.

Clearly the power rule is true for $n=1$ since the derivative of $x$ is clearly 1. So the only thing to do is to assume that the power rule is true for $n=k$. This means that we assume that we know for a fact that the derivative of $x^k$ is $k{x}^{k-1}$ and we want to prove that the derivative of $x^{k+1}$ is what it is supposed to be, which is $(k+1)x^k$.

To do so, we ask you to participate by proving the following formula yourself:

Exercise: prove the following for yourself.
The derivative of   $h(x)=xf(x)$   is   $h^{\prime }(x)=f(x)+x{f}^{\prime }(x)$

Now, back to our proof. We simply need to compute the derivative of $x^{k+1}$ and make sure we get the desired result using only things we already assume true in the first place. But since

$x^{k+1}=x\cdot x^k$

we can use the exercise above to compute the derivative of this product. We get that its derivative is

$x^k+x\cdot k{x}^{k-1}$

since we assume that the derivative of $x^k$ is $k{x}^{k-1}$. Let's simplify what we found now. We have that the derivative of $x^{k+1}$ is

$x^k+x\cdot k{x}^{k-1}=x^k+k{x}^k=(k+1)x^k$

which is exactly what we were looking for and hence finishes our proof.