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# Differentiating polynomials

AP.CALC:
FUN‑3 (EU)
,
FUN‑3.A (LO)
,
FUN‑3.A.3 (EK)

## Video transcript

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 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 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 view this as the derivative operator this says look I want to take the derivative of whatever is inside of the parenthesis 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 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 is just the derivative of is equal to the sum or the difference of D of the derivative of each of them so this is equal to the derivative let me just it's 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 to the fifth so I'll put the X 2/5 X to the fifth there here I had a 2x here I had a 2x 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 that I'm adding or subtracting them 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 4x to 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 5x we could say to the five minus one power which of course is just four and then for this second one we could 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 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 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 1/3 be well we would bring the 3 out front and decrement the exponent and so this would be equal to this 2 times the 3 times X to the 3 minus one 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 I could just write a 6x squared here so I could just so this is going to be 6x 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 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 the power rule right over here so bring the two out front and decrement the exponent so it's going to 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 us 2x so just like that we have been able to figure out the derivative of f and 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 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 going to be five times two to the fourth plus plus six times two squared six times 2 squared minus minus 2 times 2 minus two times two and this is going to be equal to let's see 2 to the 4th 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 that if you look if you were to graph the tangent line when X is equal to 2 for every ink 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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