Tutorials on Advanced Math and Computer Science Concepts

# Derivatives of Polynomials

As derivatives are an important concept, it is beneficial to understand the laws and properties that exist for them, so that we can evaluate derivatives quickly and easily. Throughout this section we will look at common classes of functions and determine rules for calculating their derivatives.

Constant Derivatives: If f(x) is a constant value, c, then f'(x) = 0.

Let's prove that this is true using the definition of a limit we found in the last section.

$lim_{h \to 0} \frac{f(x+h)-f(x)}{h} = lim_{h \to 0} \frac{c-c}{h} = lim_{h \to 0}0 = 0$

Moving forward from this, we can next look at power functions, meaning any function in the form $f(x) = x^n$

First, let's take a look at the derivative of the simplest power function, f(x) = x. Before even calculating the derivative, we know the slope is 1 at all points, since the function is linear. Let's see if the derivative is consistent with this.

$lim_{h \to 0} \frac{f(x+h)-f(x)}{h} = lim_{h \to 0} \frac{x+h-x}{h} = lim_{h \to 0} \frac{h}{h} = 1$

Therefore, as expected, $\frac{dy}{dx}(x) = 1$

Let's continue on and see if we can find a pattern. If we look at $\frac{dy}{dx}(x^2)$, we get:

$lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$
$=lim_{h \to 0} \frac{(x+h)^2-x^2}{h}$
$=lim_{h \to 0} \frac{x^2+2xh+h^2-x^2}{h}$
$=lim_{h \to 0} \frac{x^2+2xh+h^2-x^2}{h}$
$=lim_{h \to 0} \frac{h(2x+h)}{h}$
$=lim_{h \to 0} 2x+h = 2x$

So, $\frac{dy}{dx}(x^2) = 2x$. If you continued calculating derivatives, you would find that $\frac{dy}{dx}(x^3) = 3x^2$, $\frac{dy}{dx}(x^4) = 4x^3$, and so on.You should see that a pattern emerges from this. In general:

Power Rule: If $f(x) = x^n$, then $f'(x) = nx^{n-1}$

Let's see if we can prove that this is true. In order to do this, we need to know how to expand $x^n$.This task requires the binomial theorem, which we will see in the second step of the proof.

Take $f(x) = x^n$. If we take the derivative through our definition, we get:

$lim_{h \to 0} \frac{(x+h)^n-x^n}{h}$
$=lim_{h \to 0} \frac{(x^n+nx^{n-1}h+\frac{n(n-1)}{2}x^{n-2}h^2+...+nxh^{n-1}+h^n)-x^n}{h}$
$=lim_{h \to 0} \frac{nx^{n-1}+\frac{n(n-1)}{2}x^{n-2}h^2+...+nxh^{n-1}+h^n}{h}$
$=lim_{h \to 0} nx^{n-1}+\frac{n(n-1)}{2}x^{n-2}h+...+nxh^{n-2}+h^{n-1}$
$=nx^{n-1}$

This proves that the power rule is true for any $f(x) = x^n$. You should note that this works for n as any real number, meaning that this applies to negative and fraction exponents as well.

Example: Given $f(x) = \frac{1}{x^2}$, find f'(x)

To do this, recall that $\frac{1}{x^2} = x^{-2}$. From here we can apply the power law:

$\frac{dy}{dx}(x^{-2}) = -2x^{-3}$

Example: Given $f(x) = \sqrt{x^3}$, find f'(x)

To do this, recall that $\sqrt{x^3} = (x^3)^{\frac{1}{2}} = x^\frac{3}{2}$

The power rules are very valuable, as they can help us derive rules for polynomial functions, as well as others. Before we take on polynomials however, we need a few more rules.

The constant multiple rule: Suppose that c is a constant, and f(x) is a differentiable function. Then, $\frac{dy}{dx}(c[f(x)]) = c\frac{dy}{dx}(f(x))$

This simply means that when we see a constant, we can factor it out, apply the derivative to the function, then multiply the result by the constant. Let's prove that this rule is true.

$lim_{h \to 0}\frac{cf(x+h)-cf(x)}{h}$
$lim_{h \to 0}c \frac{f(x+h)-f(x)}{h}$
$c*lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$
$=c*f'(x)$

As you can see, applying limit laws allows us to factor the constant out of the limit, arriving at the result we expected in the rule. A similar result for sums and differences can be found by applying limit laws as well.

The Sum Rule: If f(x) and g(x) are both differentiable, then: $\frac{dy}{dx}(f(x)+g(x)) = \frac{dy}{dx}(f(x)) + \frac{dy}{dx}(g(x))$

To prove this, we simply need to apply limit laws for addition.

$lim_{h \to 0} \frac{f(x+h) + g(x+h)) - (f(x) + g(x))}{h}$
$=lim_{h \to 0} \frac{f(x+h)-f(x)}{h} + lim_{h \to 0} \frac{g(x+h)-g(x)}{h}$
$=f'(x)+g'(x)$

Similarly, the same logic applies to differences.

The Difference Rule: If f(x) and g(x) are both differentiable, then: $\frac{dy}{dx}(f(x)-g(x)) = \frac{dy}{dx}(f(x)) - \frac{dy}{dx}(g(x))$

The proof of this is the exact same as the sum rule, just changing addition to subtraction.

With these rules, we can now successfully find derivatives for polynomials.

Example: Given $f(x) = x^3+x^2+5x+7$, determine f'(x)

$\frac{dy}{dx}(x^3+x^2+5x+7)$
$=\frac{dy}{dx}(x^3) + \frac{dy}{dx}(x^2) + \frac{dy}{dx}(5x) + \frac{dy}{dx}(7)$
$=3x^2+2x+5* \frac{dy}{dx} (x) +0$
$=3x^2+2x+5$