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### Course: AP®︎/College Calculus AB>Unit 2

Lesson 7: Derivative rules: constant, sum, difference, and constant multiple: connecting with the power rule

# Differentiating polynomials

Let's explore how to differentiate polynomials using the power rule and derivative properties. We work with the function f(x)=x⁵+2x³-x² and apply the power rule to find its derivative, f'(x)=5x⁴+6x²-2x. Next, we evaluate f'(x) at x=2, determining that f'(2)=100, which represents the rate of change or slope of the tangent line at the specified point.

## Want to join the conversation?

• What do you do when you have a function like f(x)=4x^2+2x? I know that for the first part (4x^2) you can use the power rule, but what about the second part (2x)?
• Basically - x^1 becomes x^0, which becomes 1. Any constant times 1 becomes that constant. Therefore f'(x) of 2x = 2 * x^(1-1) = 2*1 = 2. I just like to think that x^1 "disappears" when derived, leaving just the constant in front of it.
• In this video and in others, Sal takes the derivative of both sides of the equation. Is that always “legal”, mathematically speaking?
• Yes, if two functions are equal, then their derivatives are equal.
• How do you take derivatives with respect to y?
• When Sal writes f(x), he just uses x as a variable, and he means f(a variable). Therefore, you could substitute f(x) for f(y), and get the same graph. As to actually taking the variable, you would do d*f(y)/dy, or take the limit as h approaches 0 of [f(y+h) - f(y)]/h
• How can we find minimum/maximum values of polynomials by differentiation?
• Given that with the Derivative we are able to get the Slope of tangent lines to our function at any x values, if we set our Derivative expression equal to 0 we are going to find at what x values we have the Slope of our tangent line equaling 0, which would be just a horizontal line.

The only time that happens is at min/max values. We can only get horizontal tangent lines at those values, if you think about it. With the rest of our function our tangent lines have Slopes that are either decreasing or increasing along the function.

We call these x values Critical numbers. Where we have horizontal tangent lines.

To determine whether we have a min/max, we would test other x values around the Critical points we already found into our Derivative function (evaluating it), to see if our Slope is positive (that would imply that the function is increasing on that interval), or if it is negative (our function would be decreasing).

If there's a change from having a positive Slope (our function increasing) to have a negative Slope (our function decreasing), then that is a Relative/Local maximum. If there's a change from decreasing to increasing, then that is a Relative/Local minimum.

Finally, we would evaluate our Critical numbers into our original function to actually figure out the points of our mins/maxes.

The process of doing this, finding Critical numbers by setting our Derivative equal to 0, and testing values around them to determine whether we have a min/max, is known as the First Derivative Test.
• @ Why is the derivative of polynomial is the sum of the difference of derivative of each item? It is "given" in this video, but I want to understand "why".
• Let f(x) and g(x) be two functions. From the definition of a derivative:

lim(h-->0)f(x) = f(x+h)-f(x)/h

lim(h-->0)g(x) = g(x+h)-g(x)/h

Now, add the limits. You'll get lim(h-->0)(f(x) + g(x)) = (f(x+h)-f(x)+g(x+h)-g(x))/h

Now, let a function a(x) exist such that a(x) = f(x) + g(x). So, lim(h-->0)a(x) = a(x+h)-a(x)/h

As h(x) is just f(x) + g(x), we have:

lim(h-->0)(f(x)+g(x)) = (f(x+h)-f(x)+g(x+h)-g(x))/h

This clearly shows that derivative of the sum/difference = sum/difference of the derivatives.
• What would be the answer for
2x^2 + 30x = 0
Value for x
We can calculate it as 15 by using simple equation formula. But what about derivative method
• Not totally sure about what you want to convey...The solutions to that parabola are 0 and -15 and its derivative is 4x +30 by using the power rule.....the slope of this linear function is 4.....
• that's some funky notation d/dx. you'd think the d's would cancel out.
• 'dx' is a single symbol, and refers to the differential of x. Think of it as an infinitesimally small amount of x. It is not the product of d and x.
• I learned this but I cant differentiate 2^x. How can I do it ?
• 2^x is exponential, so you can't differentiate it like a polynomial. d/dx[2^x] = ln(2) * 2^x.
• I think this means that if x is at 2, then an incremental increase in x of +1 (i.e. 2 + 1 = 3) results in an increase in y of +100. This is a Rate of Change of a slope at x=2. Is that correct?

Later added note: We also know that at x=2, y=44, and at x=3, y= 288. I find that counter-intuitive, as I would expect the incremental increase from x=2 to x=3 to be +100 i.e. the derivative of the original function. I think the answer is that the derivative (of 100) is an instantaneous rate of change AT x=2, not FROM x=2 to x=3
• Sorta, yeah!
What you're doing is approximating the change in `y` from `x=2` to `x=3`, by multiplying the derivative at `x=2` by the difference in `x`.
Then, using that approximate change in `y` to likewise approximate the value of `y` at `x=3`, by adding it from `y` when `x=2`.

This is learned as the increments formula! It can be used when you know the starting `x`, the end `x` and `f'(starting x)`.

Also your answer is exactly correct; it's only an approximate because the rate of change of the function is constantly changing within the interval (`x=2` to `x=3`).
• What do you do when you have a function like f(x)=4x^2+2x? I know that for the first part (4x^2) you can use the power rule, but what about the second part (2x)?
• You can still use the power rule on the 2x (x is still raised to the 1 power so the derivative of 2x is 2). Using the power rule you multiply the power of x which in the case of 2x is 1 and 2*1 yields 2 then subtract 1 from the power of x you get x^0 which is equal to 1. You end up with 2*1 which is 2. The derivative of 2x is 2.

## Video transcript

- [Voiceover] So I have the function F of X here and we're defining it using a polynomial expression. And what I would like to do here is take the derivative of our function which is essentially gonna make us take the derivative of this polynomial expression and we're gonna take the derivative with respect to X. So the first thing I'm gonna do is let's take the derivative of both sides. So we can say the derivative with respect to X of F of X of F of X is equal to the derivative with respect to X the derivative with respect to X of X to the fifth X to the fifth plus two plus two X to the third minus X squared. And so the notation, just to get familiar with it you could do this as the derivative operator. This is, I want to take the derivative of whatever's inside of the parentheses with respect to X. So the derivative of F with respect to X we could use the notation but that is just F prime of F prime of X. And that is going to be equal to. Now here we can use our derivative properties. The derivative of the sum or difference of a bunch of things. The derivative of is equal to the sum of the difference of the derivative of each of them. So this is equal to the derivative let me just, with the derivative with respect to X of each of these three things. So the derivative with respect to X. Let me just write it out like this. Of that first term plus the derivative with respect to X of that second term minus the derivative with respect to X of that third term. Of that third term. And I'll color code it here. So here I had an X of the fifth so I'll put the X to fifth X of the fifth there. Here I had a two X to the third. So I'll put the two X to the third there. And here I have a X squared I'm subtracting an X squared so I'm subtracting the derivative with respect to X of X squared. So notice all that's happening here is I'm taking the derivative individually of each of these terms and then I'm adding or subtracting in the same way that the terms were added or subtracted. And so what is this going to be equal to? Well, this is going to be equal to for X of the fifth we can just use the power rule. We can bring the five out front and decrement the exponent by one so it becomes five X we can say to the five minus one power. Which of course, is just four. And then for this second one we could do it in a few steps. Actually, let me just write it out here. So I could write I could write the derivative with respect to X of two X to the third power is the same thing it's equal to the same, we could bring the constant out. The derivative with, two times the derivative with respect to X of X to the third power. This is one of our this is one of our derivative properties. The derivative of a constant times some expression is the same thing as a constant times the derivative of that expression. And what will the derivative with respect to X to the third be? Well, we would bring the three out front and decrement the exponent. And so this would be equal to this two times the three times X to the three minus one power. Which is, of course, the second power. So this would give us six X squared. So, another way that you could have done it I could just write I could just write a six X squared here. So I could just, so this is going to be six X squared. And, instead of going through all of this you'll learn as you do more of these, that you could have done this pretty much in your head. Say look, I have the three out here as an exponent. Let me multiply the three times this coefficient because that's what we ended up doing anyway. Three times the coefficient is six X and then three minus one is two. So you didn't necessarily have to do this but it's nice to see that this comes out of the derivative properties that we talk about in other videos. And then finally, we have minus. And we use the power rule right over here. So, bring the two out front and decrement the exponent. So it's gonna be two. It's going to be two times X to the two minus one power which is just one. Which, we could just write as two X. So just like that we have been able to figure out the derivative of F. And you might say, well what this thing now? Well now we have an expression that tells us the slope of the tangent line. Or you could view it as the instantaneous rate of change with respect to X for any X value. So, if I were to say if I were to now to say F prime, let's say F prime of two. This would tell me what is the slope of the tangent line of our function when X is equal to two. And I do that by using this expression. So this is gonna be five times two to the fourth plus plus six times two squared. Six times two squared. Minus minus two times two. Minus two times two. And this is going to be equal to let's see, two to the fourth power is 16, 16 times five is 80. And that's 80, and then this is six times four, which is 24. And then we are going to subtract four. So this is 80 plus 24 is 104 minus four is equal to 100. So, when X is equal to two this curve is really steep. The slope is 100. If you look, if you were to graph the tangent line when X is equal to two for every positive movement in the X-direction by one you're gonna move up in the Y-direction by 100. So it's really steep there and it makes sense. This is a pretty high degree. X to the fifth power and then we're adding that to another high degree X to the third power. And then we're subtracting a lower degree. So that's what you would expect.