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*.

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)*=3*x*, 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 2*x* + *delta x*. Well, as *delta x* goes to zero, this is just equal to 2*x*, so the derivative of *f(x)*=*x*^2 is equal to *f'(x)*=2*x*. When *x*= 0, *f(x)*=0 and *f`(x)*= 0. At *x*=2, *f(x)*= 4 and *f`(x)*=2*x*=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 3*x*^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 2*x*. When *y*=*x*^3, the derivative is 3*x*^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 0*x*^(0 – 1), which is just 0. - When
*y*=*x*^1, we had*n*=1, so our derivative was 1*x*^(1 – 1), or 1*x*^0, which is just 1. - When
*y*=*x*^2, we had*n*=2, so our derivative was 2*x*^(2 – 1), or just 2*x*. - When
*y*=*x*^3, we had*n*=3, so our derivative was 3*x*^(3 – 1), or just 3*x*^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)*=47*x*^(47 – 1), which simplifies as 47*x*^46. You can even use that formula when *f(x)*=*x*^-2. In this case, your *n* is -2, so your derivative is -2*x*^(-2 – 1) or just -2*x*^-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)*=-16*t*^2 + 36*t*. 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*(-16*t*^2 + 36*t*)- We can divide and conquer for
*h`(t)*=*d/dt*(-16*t*^2) +*d/dt*(36*t*) - We can pull out the constants for
*h`(t)*= -16*d/dt*(*t*^2) + 36*d/dt*(*t*) - Using the power rule for derivatives, we have
*h`(t)*= -16(2*t*) + 36(1) - We can expand that out for
*h`(t)*= -32*t*+ 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).