Polynomials can describe just about anything and are especially common in describing motion. Learn the tricks to quickly finding the derivatives of these ubiquitous functions.

## Basic Rules Let’s take a look at Super C, the human cannonball, and let’s look at his height as a function of time. We know that we can find his vertical velocity, or how much his height changes as a function of time, by taking the derivative of his height. We can take this limit formally by using the formula h'(t) = the limit as delta t approaches zero of (h(t + delta t) – h(t))/delta t.

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But there’s an easier way, so let’s talk about taking the limits of powers and polynomials. These are powers, like f(x)=x^6, and polynomials, like f(x)= 3 + x^47 – x^22 + (1/2)x^15, and so on and so forth. We know that, in the case of a polynomial, we can divide and conquer. We can look at each term on its own. We just need to find the derivative of 3, then the derivative of x^47 and so on and so forth.

Let’s start with the easiest of derivatives. Let’s start with a constant, like f(x)=1. If I graph this, I have a straight line. Furthermore, I know that the derivative is the slope of the tangent of this line, and I could tell you what that derivative is just by looking at this graph. The slope of that line is constant; it’s always equal to zero. So the derivative of some constant, like f(x)=1, is equal to zero.

Let’s look at a slightly more complicated one, like f(x)=x. If I graph this, then I see again that the slope is constant. Further, I know that saying f(x)=x is like saying y=x + 0. From slope-intercept form, I know that the slope here is 1, and I can calculate that anywhere along the line. If f(x)=3x, I know that the slope is 3, so the derivative is going to be equal to 3. ## Finding Derivatives of Polynomials

Okay, this isn’t so bad. What about a case like f(x)=x^2? When graphing this, the slope of the tangent is not always constant. So when finding the derivative of x^2, I need to use a formal calculation. Let’s use f`(x)= the limit as delta x approaches zero of (f(x + delta x) – f(x))/delta x. Let’s plug in f(x + delta x), which is (x + delta x)^2, and f(x), which is x^2. We can expand the (x + delta x)^2 and solve to see that the limit as delta x goes to zero is 2x + delta x. Well, as delta x goes to zero, this is just equal to 2x, so the derivative of f(x)=x^2 is equal to f'(x)=2x. When x= 0, f(x)=0 and f`(x)= 0. At x=2, f(x)= 4 and f`(x)=2x=4. So the slope of the tangent at that point equals 4.

Let’s look at one more. Let’s look at f(x)=x^3. Now the derivative still equals (f(x + delta x) – f(x))/delta x. But now f(x + delta x) = (x + delta x)^3 and f(x)=x^3, so this gets more complicated. Once again, you can expand out the (x + delta x)^3, simplify the terms and divide the top and bottom by delta x. You find that the derivative of x^3 is 3x^2.

## The Power Rule for Derivatives

We could keep going, but let’s stop and take a look at the pattern that we have. Let’s say y=f(x) for each of these cases, so when y=1, then y`=0. What I could say instead of y=1, is y=x^0, since anything to the zero power equals 1. When y=x, I can write that as x^1, the derivative was 1. When y=x^2, the derivative is 2x. When y=x^3, the derivative is 3x^2. So you can see there’s a pattern going on here, and this leads to a good power rule for derivatives. If f(x)=x^n, then the derivative f`(x)=nx^(n – 1).

• When we had y=x^0, n=0, so our derivative was 0x^(0 – 1), which is just 0.
• When y=x^1, we had n=1, so our derivative was 1x^(1 – 1), or 1x^0, which is just 1.
• When y=x^2, we had n=2, so our derivative was 2x^(2 – 1), or just 2x.
• When y=x^3, we had n=3, so our derivative was 3x^(3 – 1), or just 3x^2.

You can do this for anything, like f(x)=x^47. You could expand that out to the 47th power, but using our rule, you could just write f`(x)=47x^(47 – 1), which simplifies as 47x^46. You can even use that formula when f(x)=x^-2. In this case, your n is -2, so your derivative is -2x^(-2 – 1) or just -2x^-3.

## Calculating Vertical Velocity

Now let’s go back and look at the vertical velocity of Super C the human cannonball. Again, we can look at his height as a function of time is h(t)=-16t^2 + 36t. Now you can go back and do this formally, or you can just use our new rule:

• h`(t) = d/dt(h(t)) = d/dt(-16t^2 + 36t)
• We can divide and conquer for h`(t) = d/dt(-16t^2) + d/dt(36t)
• We can pull out the constants for h`(t) = -16d/dt(t^2) + 36d/dt(t)
• Using the power rule for derivatives, we have h`(t) = -16(2t) + 36(1)
• We can expand that out for h`(t) = -32t + 36. This is the same answer that we would’ve gotten if we’d done the formal and much more complicated expansion of the derivative.

## Lesson Summary

When you’re finding the derivatives of any kind of power or polynomial, always remember the quick rule: If we have a function f(x)=x^n, then the derivative, f`(x)=nx^(n – 1).