Calculating Derivatives of Polynomial Equations

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  • 0:05 Basic Rules
  • 2:24 Finding Deriviatives…
  • 4:50 Power Rule for Derivatives
  • 8:03 Calculating Vertical Velocity
  • 10:02 Summary
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Lesson Transcript
Instructor: Robert Egan
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

The slope here is 1 and can be calculated anywhere along the line
Polynomial Derivatives Graph Example 2

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.

The slope of the tangent equals 4
Polynomial Derivatives Graph Example 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.

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