This article will walk you through how to prove the Power Rule and take the derivative of any polynomial. We are going to prove every step of the way. If you don’t want to see the proof and just want a formula, then just know that for any function $f(x) = x^n$, $f'(x) = nx^{n-1}$; this is called the Power Rule. If you want to know how this is derived, read on.

**The Product Rule**

It might seem strange, but to prove the Power Rule, we first need to prove the Product Rule (there are other ways to prove it, but this way is probably easiest to understand). When we use the Product Rule, we want to find the derivative of two differentiable functions multiplied together, $(f(x)\cdot g(x))’$. Let’s use the definition of a derivative:

\[ \begin{align*} (f(x)\cdot g(x))’ = \lim_{h\to 0} \dfrac{f(x+h)g(x+h) – f(x)g(x)}{h} \end{align*} \]

Let us do a clever manipulation by adding $-f(x+h)g(x)+f(x+h)g(x)$ to the numerator. This doesn’t change the value of the limit because we’re adding 0, but it will allow us to evaluate the limit:

\[ \begin{align*} &\lim_{h\to 0} \dfrac{f(x+h)g(x+h) – f(x+h)g(x) + f(x+h)g(x) – f(x)g(x)}{h} &=\lim_{h\to 0} \dfrac{f(x+h)(g(x+h)-g(x)) + g(x)(f(x+h)-f(x)}{h} &=\lim_{h\to 0} \dfrac{f(x+h)(g(x+h)-g(x))}{h} + \lim_{h\to 0} \dfrac{g(x)(f(x+h)-f(x))}{h} &=(\lim_{h\to 0} f(x+h))(\lim_{h\to 0} \dfrac{g(x+h)-g(x)}{h}) + (\lim_{h\to 0} g(x))(\lim_{h\to 0} \dfrac{f(x+h)-f(x)}{h}) \end{align*} \]

Since both $f(x)$ and $g(x)$ are differentiable $\displaystyle \lim_{h\to 0} \dfrac{f(x+h) – f(x)}{h} = f'(x)$ and $\lim_{h\to 0} \dfrac{g(x+h) – g(x)}{h}= g'(x)$, by definition. Additionally, $\displaystyle \lim_{h\to 0} f(x+h)=f(x)$, and $\displaystyle \lim_{h\to 0} g(x)=g(x)$. We may thus rewrite the last line as $f(x) g'(x) + f'(x) g(x)$, which is the Product Rule:

$(f(x)\cdot g(x))’$ = $f'(x) g(x) + f(x) g'(x)$.

**Proving The Power Rule**

We can now use the Product Rule to prove the Power Rule, which says that the derivative of any function $f(x) = x^n$ is $nx^{n-1}$, where $n$ is a constant. It is very important to remember that the Power Rule applies only when the exponent is a constant. Many people improperly apply the Power Rule to functions like $f(x) = x^{2x}$, which does not work because $2x$ is not a constant. Our proof goes as follows:

We can use the definition of a derivative to prove that the derivative of $x$ is $1$: $\lim_{h\to 0} \dfrac{x + h – x}{h} = \lim_{h\to 0} \dfrac{h}{h} = 1$.

Let us now find the derivative of $x^2$: $x^2 = x \cdot x$, so we can use the Product Rule:

$\frac{d}{dx}(x^2) = \frac{d}{dx}(x\cdot x) = 1\cdot x + x\cdot 1 = 2x$.

We can now find the derivative of $x^3$:

$\dfrac{d}{dx}(x^3) = \frac{d}{dx}(x^2\cdot x) = 2x\cdot x + x^2\cdot 1 = 3x^2$.

In general this process will look like this:

$\dfrac{d}{dx}(x^n) = \frac{d}{dx}(x^{n-1}\cdot x)$ = $(n-1)x^{n-2}\cdot x + x^{n-1}\cdot 1 = nx^{n-1}$.

This proof only works when $n$ is a positive integer. Proving this for any $n$ requires knowledge of more advanced concepts such as implicit differentiation.

**Proof by Implicit Differentiation**

This proof is quite a bit shorter and does not involve the Product Rule, but it does require implicit differentiation and a bit of mathematical trickery.

Let there be some function $y = x^n$, where $n$ is a constant. We can take the logarithm of both sides to get $\ln y = \ln x^n = n\ln x$. Implicitly differentiating, we get that $\dfrac{y’}{y} = \dfrac{n}{x}$, and thus $y’ = \dfrac{ny}{x} = \dfrac{nx^n}{x} = nx^{n-1}$. This proof works for all $n$.

**Conclusion**

Now you have learned how to take the derivative of any polynomial. You also know how to prove the Power Rule in multiple ways. Good luck on your future math adventures!