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.
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.
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.
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).