AP Calculus AB — Product and Quotient Rules
Corrected lesson transcript


## 0:00 — Product and Quotient Rules

[0:05] Okay, before we get started, what functions do we now know the derivative of very easily?
[0:27] Any polynomial?
[0:39] And why do we know the derivative of any polynomial?
[0:46] Power rule and?
[0:52] Sum and difference rule.
[0:53] So, the polynomial is just a sum of multiple power functions.
[1:00] And the sum and difference rules we separate by addition, we take the derivative separately.
[1:07] Just for funsies, what's the derivative of this?
[1:11] The function is x to the fifth plus 3x squared minus 7x plus 2.
[1:14] Its derivative is 5x to the fourth plus 6x minus 7.
[1:17] The derivative of the constant term is zero.
[1:19] ok, what other functions?
[1:23] Uh, okay, sorry, what else?
[1:27] What's that?
[1:28] Square root functions.
[1:29] Square root functions, that's true.
[1:32] Any kind of power function.
[1:37] Write the cube root of x, which is x to the one third,
[1:41] which is a power function the derivative is 1 3rd minus 1 is negative 2 3rds
[1:49] sounds good what other functions sine and cosine two of the trig functions
[1:56] derivative of sine is cosine and derivative of cosine is negative sine and then one
[2:06] What is the one other function? e to the x. What is its derivative? e to the x.
[2:32] So here's an example of the product of two functions.
[2:41] It's not the sum of it.
[2:44] If it were the sum of x to the fourth and sine x, we could differentiate each term.
[2:46] we could easily do it.
[2:48] But we have a product and it turns out it is not,
[2:52] the derivative, don't write this down.
[2:55] It is not 4x cubed times cosine x.

## 2:58 — Why Products Need a New Rule

[3:00] not the derivative of one function
[3:02] times the derivative of the other.
[3:04] And let me just give you an example.
[3:13] X to the fifth, same thing as X squared times X cubed.
[3:19] We agree?
[3:20] Yeah.
[3:21] We know the derivative of X to the fifth,
[3:25] which is what, five X to the fourth.
[3:30] Derivative of X squared is two X,
[3:33] derivative of this is three X squared,
[3:36] which is six X cubed, which is definitely not this, right?
[3:41] There is a rule called the product rule. If y equals f of x times g of x,

## 3:48 — State the Product Rule

[3:55] then its derivative has two terms.
[4:06] It turns out it's this. The derivative of the first function
[4:12] times the second plus the first function
[4:17] times the derivative of the second function.
[4:53] Apply the product rule to x squared times x cubed.
[4:55] The derivative of the first function is 2x, times the second function,
[5:06] plus the first function times the derivative of the second function.
[5:19] This is 2x to the fourth plus 3x to the fourth, which equals 5x to the fourth.
[5:34] showing an example that works out is not proof that's always true how do we know
[5:41] this is always true let's prove it and let me call this a H of X just for a

## 6:03 — Prove the Product Rule

[6:05] Let m of x equal f of x times g of x.
[6:19] And how are we going to prove it?
[6:24] The definition of the derivative.
[6:28] m prime of x is the limit as h approaches zero of m of x plus h minus m of x, all divided by h.
[6:37] I shouldn't use h's here, but that's okay.
[6:45] Smaller h. Aye yi yi, too many h's. Let's call this something else.
[6:54] Call the product m of x.
[7:15] For m of x plus h, substitute x plus h into both factors.
[7:19] X.
[7:30] The numerator is f of x plus h times g of x plus h minus f of x times g of x, all divided by h.
[8:00] Now what?
[8:04] Any ideas?
[8:14] No ideas?
[8:19] Should we try multiplying by the conjugate? No, it's not gonna help. There is a trick.
[9:03] All I do is separate those two terms.
[9:12] We're going to add something here.
[9:15] Anybody want to take a guess what we're going to add?
[9:21] The only thing we're allowed to do is add zero.
[9:25] It doesn't change anything, right?
[9:27] We're going to add something and then we're going to subtract something.
[9:32] H plus h.
[9:35] I like your idea, but I don't think that's going to work.
[9:38] We're going to add something so then we could factor out something in common.
[9:50] We're going to add negative.
[10:27] Let me think for a second what it is.
[10:47] So we're going to add...
[10:52] Let's try this.
[10:52] It might not work out.
[10:55] I think this is it.
[10:56] So if we add a negative f of x, g of x plus h, we're going to have to add a positive f
[11:04] of x times g of x plus h.
[11:11] All right, so all I did was add the opposites, which is just 0.
[11:24] Now what can I do with these first two terms?
[11:33] What can I factor out?
[11:34] g of x plus h.
[11:42] And I'm going to change this into two fractions.
[11:46] This and this.
[11:48] So g of x plus h.
[12:10] Let me change it to two different limits.
[12:14] From the second pair of terms, what can I factor out?
[12:16] F of X.
[12:36] And we're very close now.
[12:46] This product here, I'm going to change it to two different limits.
[13:50] And what is this guy right here?
[13:56] That is the definition of the derivative of f of x.
[14:02] That's f prime of x.
[14:07] And as h goes to 0 here, this goes to g of x.
[14:17] There is no H's, so this is just F of X.
[14:25] And this right here is the definition of the derivative of G.
[14:33] And let me just switch the order
[14:44] And there it is we just proved the product rule
[14:51] The first person to prove this was clever; that first move is tricky.

## 14:53 — Apply the Product Rule

[15:00] But there it is.
[15:02] Whenever we differentiate a product of two functions, we use the product rule.
[15:08] We have to use the product rule.
[15:10] Let's do some examples.
[15:17] Oh actually, let's do that example I had up here.
[15:20] What was it?
[15:21] x to the fourth times sine x.
[15:42] So the derivative
[15:56] So it's the derivative of our first function times our second function plus the derivative,
[16:04] sorry, first function times the derivative of the second function.
[16:19] So we'll just write this down.
[16:31] And the derivative of x to the fourth is?
[16:33] 4x cubed.
[16:34] 4x cubed times sine x plus x to the fourth times cosine x.
[16:42] Derivative of sine of x?
[16:44] Cosine x.
[16:48] And that's it.
[16:56] Easy peasy.
[16:59] Please do it in this order where it's the derivative of the first function times the
[17:03] second plus the first times the derivative of the second.
[17:07] Either way, I like it this way. It's going to be very important for the quotient rule.

## 17:15 — Products with Three Factors

[17:17] Okay, let's do another one.
[17:41] Uh oh, we got a problem here.
[17:44] What's the problem?
[17:46] It's three terms.
[17:48] It's not three terms.
[17:51] Terms are things separated by addition.
[17:53] There are three...
[17:55] Or it's a product of three things.
[17:58] There are three factors.
[18:00] Okay, it's not just a product of two functions.
[18:02] It's the product of three functions.
[18:04] So now what? What's the idea?
[18:16] Okay, I like that.
[18:20] So let's, which should we pair up? Let's call these two.
[18:24] One function.
[18:29] We're going to use the product rule right here.
[18:35] So it's the derivative of the first times the second plus the first times the derivative
[18:44] of all the second
[19:17] x to the fifth times cosine x times e to the x.
[19:26] Uh oh, then what are we doing when we get to here?
[19:29] Yeah, we got, again, we're finding the derivative of a product, so we got another product rule.
[19:39] Oops.
[20:05] Deriv of cosine is negative sine.
[20:16] Deriv of e to the x is e to the x.
[20:24] And then let me distribute this.
[20:50] There it is.
[20:55] We just gotta use the product rule twice.
[21:02] Is there any shortcut?
[21:08] Instead of using the product rule once and using it twice?
[21:20] So tell me what to do.
[21:22] First, differentiate the first function and multiply by the other two.
[21:27] That gives 5x to the fourth times cosine x times e to the x.
[21:29] And it would be 5x to the fourth times cosine times x.
[21:35] Then differentiate the second function and multiply by the other two.
[21:39] And then what?
[21:40] Differentiate cosine x.
[21:43] This gives x to the fifth times negative sine x times e to the x.
[21:49] times x to the x.
[21:53] And then?
[21:55] Finally, differentiate the third function.
[21:57] This gives x to the fifth times cosine x times e to the x.
[22:06] For three factors, differentiate one factor at a time and multiply by the other two, then add all three terms.
[22:18] And that is exactly what we have here, derivative of the first function times the other two,
[22:31] plus the derivative of the second function times the other two, plus the derivative of
[22:39] The derivative of the third function may look unchanged because e to the x is its own derivative.
[22:46] always gonna work? Definitely. Okay we could do it with f of x, g of x, and h of
[22:58] x. Just use the product rule without these actual functions. It will always work. So if we have four
[23:05] functions let's say this was a times sine of X or something we just have what
[23:15] we have to have sine of X in here sine of X sine of X and then plus the first
[23:24] three functions times the derivative of sine of X okay so we can have a product
[23:30] to more than just two it's just the derivative of the first function times the other two
[23:36] plus the derivative of the second function times the other two and so on. Okay not too bad.
[24:14] Can we use the product rule here?
[24:19] Could we change this into a product?
[24:22] I could change this into a product.
[24:25] That's x to the cubed times what?
[24:26] One over sine x.
[24:29] true, 1 over sine of x, sine of x to the negative 1 power.
[24:38] We know the derivative of this, do we know the derivative of sine of x to the negative
[24:41] 1 power?
[24:45] We don't.
[24:49] We know the derivative of sine of x, but not sine of x to the negative 1.
[24:52] This is actually the composition of two functions.
[24:55] This is sine of x inside this function or...

## 25:02 — The Quotient Rule

[25:03] Okay, it's a function within a function. We need it, so yeah, so that's not gonna work.
[25:12] So what are we gonna do here? There's another rule called the quotient rule.
[25:25] let me just tell you what it is we have a quotient of two functions turns out
[25:50] the derivative. The numerator is exactly like the product rule, f prime times second function,
[26:06] but it's really the denominator function, except we have a minus here instead of a plus.
[26:18] And then the denominator turns out to be g of x squared.
[26:32] The derivative of a quotient resembles the product rule in the numerator,
[26:39] except it uses subtraction instead of addition, and the denominator is squared.
[26:55] To prove it, we would start with the definition of the derivative.

## 27:02 — Apply the Quotient Rule

[27:06] And if you can prove it by the end of the class, we'll have time at the end of the class.
[27:12] You get extra credit.
[27:14] I rarely ever offer extra credit.
[27:33] Is it going to be similar to the proof of the product rule?
[27:36] It's going to be very similar.
[27:39] We would add and subtract a carefully chosen expression.
[27:42] But then it's a little bit more complicated, now we got fractions.
[27:47] But let's not do it right now.
[27:50] We'll have plenty of time at the end of class.
[27:52] Let's just use it.
[27:56] So f of x, what was it? x cubed over sine of x?
[28:19] I
[28:19] Often just set up parentheses like that
[28:23] You can do it in your head if you want and just go right to the answer
[28:26] but it's the derivative of the numerator, the first, I call it the first function,
[28:32] times the denominator minus
[28:38] first function times the derivative of the second,
[28:42] just like the product rule but with a subtraction, and then we simply divide by the denominator
[28:49] squared.
[28:58] The derivative of x cubed is 3x squared by the power rule.
[29:14] This gives 3x squared sine x minus x cubed cosine x, all over sine squared x.
[29:30] That's it.
[29:31] Easy peasy.
[29:33] Okay.
[29:37] Could we simplify this at all?
[29:40] I mean we could manipulate it, have two different fractions, but if there was like an x squared
[29:49] here or something like that, then we could possibly simplify some x's, but we're done
[29:55] here.
[30:07] Okay.

## 30:12 — Derive the Tangent Rule

[30:18] We talked about this one last week, something like that.
[30:22] And we weren't able to find the derivative
[30:29] But how can I rewrite tangent
[30:33] Sine over cosine and now we have a quotient rule
[30:39] So we're gonna find the derivative by the quotient rule
[31:54] The derivative of sine is cosine.
[32:09] The derivative of cosine is negative sine.
[32:45] Can we simplify this at all?
[32:49] How can we simplify this?
[32:53] Can I do this?
[32:56] Definitely not!
[32:57] You go to math jail, you do that.
[32:59] You can't simplify one term with a factor.
[33:04] It has to be a common factor.
[33:08] But this simplifies to what?
[33:11] Sine squared plus cosine squared equals one.
[33:13] The Pythagorean identity.
[33:22] That is the derivative.
[33:26] However, what is one over cosine?
[33:35] Is secant.
[33:42] And so we normally write it as secant squared.
[34:00] We do want to just memorize this from now on.

## 34:02 — Derive the Secant Rule

[34:03] The derivative of tangent is secant squared. Now we know the derivatives of sine, cosine, and tangent.
[34:26] Let's do one more and we'll start our homework.
[34:37] Let's find the derivative of secant x.
[34:43] How am I going to start this?
[34:44] Rewrite secant x as one over cosine x.
[34:47] Change it to 1 over cosine.
[34:52] And what rule are we going to use?
[34:54] Quotient rule.
[34:55] Quotient rule.
[35:08] Go ahead and see if you can do this one on your own.
[36:16] Zero times cosine x minus one times negative sine x, over cosine squared x, gives sine x over cosine squared x.
[36:26] That is the derivative of secant x.
[36:29] However, we normally don't write it like that.
[36:34] we normally change this to...
[36:50] to this...
[36:54] and then what is one over cosine?
[36:57] that is secant...
[37:00] And sine over cosine is tangent
[37:06] Okay, that's how we normally write it that's the one we just memorize from now on
[37:19] Derivative of secant
[37:31] There's two more we're gonna do next week.
[37:36] Cotangent and cosecant are next; then we will have all six trigonometric derivatives.
[37:42] Okay, any questions on this?
[37:44] We're gonna stop here.