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Differentiating polynomials example | Derivative rules | AP Calculus AB | Khan Academy


4m read
·Nov 11, 2024

So I have the function f of X here, and we're defining it using a polynomial expression. What I would like to do here is take the derivative of our function, which is essentially going to make us take a derivative of this polynomial expression, and we're going to take the derivative with respect to X.

So the first thing I'm going to do is let's take the derivative of both sides. So we could say the derivative with respect to X of f of X is equal to the derivative with respect to X of x to the fifth plus two x to the third minus x squared.

And so the notation, just to get familiar with it, you could view this as the derivative operator. This says, "Look, I want to take the derivative of whatever is inside of the parentheses with respect to X." So the derivative of f with respect to X, we could use the notation that that is just f prime of X.

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 is just the derivative that is equal to the sum or the difference of the derivative of each of them. So this is equal to the derivative, 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.

I'll color code it here. So here I had an x to the fifth, so I'll put the x to the fifth there. Here I had two x to the third, so I'll put the two x to the third there. And here I have an 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 that 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 5x to the fourth. We can just use the power rule; we can bring the five out front and decrement the exponent by one, so it becomes 5x to the four power.

Now, for this second one, we could do it in a few steps. Actually, let me just write it out here. I could write the derivative with respect to X of 2x 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 derivative properties: the derivative of a constant times some expression is the same thing as the constant times the derivative of that expression. And what will the derivative with respect to X of x to the third be? Well, we would bring the three out front and decrement the exponent, and so this would be equal to 2 times 3 times x to the 2 power, which is of course the second power, so this would give us 6x squared.

So another way that you could have done it, I could just write a 6x squared here. So this is going to be 6x squared. 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, saying, "Look, I have the 3 out here as an exponent; let me multiply the 3 times this coefficient," because that's what we ended up doing anyway.

3 times the coefficient is 6x, and then 3 minus 1 is 2, 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 talked 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 going to be two times x to the one power, which we could just write as 2x. So just like that, we have been able to figure out the derivative of f.

You might say, "Well, what is 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 2, 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 going to be 5 times 2 to the fourth plus 6 times 2 squared minus 2 times 2. And this is going to be equal to, let's see, 2 to the fourth power is 16; 16 times 5 is 80, so that's 80.

And then this is 6 times 4, which is 24, and then we are going to subtract 4. So this is 80 plus 24 is 104 minus 4 is equal to 100.

So when X is equal to 2, this curve is really steep; the slope is 100. If you were to graph the tangent line when X is equal to 2, for every positive movement in the X direction by 1, you're going to move up in the Y direction by a hundred. 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 it's what you would expect.

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