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Okay, before we get started, what functions do we now know the derivative of very

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easily?

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Any polynomial?

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And why do we know the derivative of any polynomial?

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Power rule and?

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Sum and difference rule.

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So, the polynomial is just a sum of multiple power functions.

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And the sum and difference rules we separate by addition, we take the derivative

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separately.

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Just for funsies, what's the derivative of this?

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The function is x to the fifth plus 3x squared minus 7x plus 2.

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Its derivative is 5x to the fourth plus 6x minus 7.

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The derivative of the constant term is zero.

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ok, what other functions?

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Uh, okay, sorry, what else?

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What's that?

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Square root functions.

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Square root functions, that's true.

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Any kind of power function.

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Write the cube root of x, which is x to the one third,

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which is a power function the derivative is 1 3rd minus 1 is negative 2 3rds

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sounds good what other functions sine and cosine two of the trig functions

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derivative of sine is cosine and derivative of cosine is negative sine and then one

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What is the one other function? e to the x. What is its derivative? e to the x.

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So here's an example of the product of two functions.

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It's not the sum of it.

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If it were the sum of x to the fourth and sine x, we could differentiate each term.

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we could easily do it.

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But we have a product and it turns out it is not,

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the derivative, don't write this down.

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It is not 4x cubed times cosine x.

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not the derivative of one function

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times the derivative of the other.

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And let me just give you an example.

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X to the fifth, same thing as X squared times X cubed.

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We agree?

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Yeah.

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We know the derivative of X to the fifth,

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which is what, five X to the fourth.

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Derivative of X squared is two X,

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derivative of this is three X squared,

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which is six X cubed, which is definitely not this, right?

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There is a rule called the product rule. If y equals f of x times g of x,

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then its derivative has two terms.

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It turns out it's this. The derivative of the first function

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times the second plus the first function

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times the derivative of the second function.

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Apply the product rule to x squared times x cubed.

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The derivative of the first function is 2x, times the second function,

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plus the first function times the derivative of the second function.

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This is 2x to the fourth plus 3x to the fourth, which equals 5x to the fourth.

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showing an example that works out is not proof that's always true how do we know

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this is always true let's prove it and let me call this a H of X just for a

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Let m of x equal f of x times g of x.

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And how are we going to prove it?

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The definition of the derivative.

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m prime of x is the limit as h approaches zero of m of x plus h minus m of x, all

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divided by h.

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I shouldn't use h's here, but that's okay.

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Smaller h. Aye yi yi, too many h's. Let's call this something else.

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Call the product m of x.

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For m of x plus h, substitute x plus h into both factors.

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X.

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The numerator is f of x plus h times g of x plus h minus f of x times g of x, all

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divided by h.

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Now what?

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Any ideas?

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No ideas?

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Should we try multiplying by the conjugate? No, it's not gonna help. There is a

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trick.

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All I do is separate those two terms.

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We're going to add something here.

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Anybody want to take a guess what we're going to add?

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The only thing we're allowed to do is add zero.

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It doesn't change anything, right?

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We're going to add something and then we're going to subtract something.

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H plus h.

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I like your idea, but I don't think that's going to work.

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We're going to add something so then we could factor out something in common.

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We're going to add negative.

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Let me think for a second what it is.

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So we're going to add...

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Let's try this.

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It might not work out.

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I think this is it.

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So if we add a negative f of x, g of x plus h, we're going to have to add a positive

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f

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of x times g of x plus h.

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All right, so all I did was add the opposites, which is just 0.

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Now what can I do with these first two terms?

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What can I factor out?

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g of x plus h.

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And I'm going to change this into two fractions.

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This and this.

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So g of x plus h.

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Let me change it to two different limits.

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From the second pair of terms, what can I factor out?

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F of X.

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And we're very close now.

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This product here, I'm going to change it to two different limits.

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And what is this guy right here?

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That is the definition of the derivative of f of x.

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That's f prime of x.

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And as h goes to 0 here, this goes to g of x.

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There is no H's, so this is just F of X.

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And this right here is the definition of the derivative of G.

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And let me just switch the order

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And there it is we just proved the product rule

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The first person to prove this was clever; that first move is tricky.

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But there it is.

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Whenever we differentiate a product of two functions, we use the product rule.

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We have to use the product rule.

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Let's do some examples.

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Oh actually, let's do that example I had up here.

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What was it?

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x to the fourth times sine x.

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So the derivative

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So it's the derivative of our first function times our second function plus the

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derivative,

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sorry, first function times the derivative of the second function.

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So we'll just write this down.

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And the derivative of x to the fourth is?

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4x cubed.

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4x cubed times sine x plus x to the fourth times cosine x.

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Derivative of sine of x?

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Cosine x.

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And that's it.

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Easy peasy.

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Please do it in this order where it's the derivative of the first function times the

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second plus the first times the derivative of the second.

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Either way, I like it this way. It's going to be very important for the quotient

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rule.

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Okay, let's do another one.

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Uh oh, we got a problem here.

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What's the problem?

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It's three terms.

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It's not three terms.

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Terms are things separated by addition.

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There are three...

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Or it's a product of three things.

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There are three factors.

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Okay, it's not just a product of two functions.

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It's the product of three functions.

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So now what? What's the idea?

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Okay, I like that.

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So let's, which should we pair up? Let's call these two.

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One function.

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We're going to use the product rule right here.

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So it's the derivative of the first times the second plus the first times the

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derivative

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of all the second

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x to the fifth times cosine x times e to the x.

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Uh oh, then what are we doing when we get to here?

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Yeah, we got, again, we're finding the derivative of a product, so we got another

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product rule.

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Oops.

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Deriv of cosine is negative sine.

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Deriv of e to the x is e to the x.

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And then let me distribute this.

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There it is.

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We just gotta use the product rule twice.

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Is there any shortcut?

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Instead of using the product rule once and using it twice?

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So tell me what to do.

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First, differentiate the first function and multiply by the other two.

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That gives 5x to the fourth times cosine x times e to the x.

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And it would be 5x to the fourth times cosine times x.

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Then differentiate the second function and multiply by the other two.

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And then what?

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Differentiate cosine x.

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This gives x to the fifth times negative sine x times e to the x.

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times x to the x.

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And then?

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Finally, differentiate the third function.

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This gives x to the fifth times cosine x times e to the x.

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For three factors, differentiate one factor at a time and multiply by the other two,

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then add all three terms.

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And that is exactly what we have here, derivative of the first function times the

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other two,

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plus the derivative of the second function times the other two, plus the derivative

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of

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The derivative of the third function may look unchanged because e to the x is its

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own derivative.

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always gonna work? Definitely. Okay we could do it with f of x, g of x, and h of

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x. Just use the product rule without these actual functions. It will always work. So

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if we have four

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functions let's say this was a times sine of X or something we just have what

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we have to have sine of X in here sine of X sine of X and then plus the first

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three functions times the derivative of sine of X okay so we can have a product

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to more than just two it's just the derivative of the first function times the other

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two

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plus the derivative of the second function times the other two and so on. Okay not

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too bad.

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Can we use the product rule here?

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Could we change this into a product?

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I could change this into a product.

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That's x to the cubed times what?

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One over sine x.

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true, 1 over sine of x, sine of x to the negative 1 power.

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We know the derivative of this, do we know the derivative of sine of x to the

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negative

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1 power?

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We don't.

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We know the derivative of sine of x, but not sine of x to the negative 1.

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This is actually the composition of two functions.

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This is sine of x inside this function or...

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Okay, it's a function within a function. We need it, so yeah, so that's not gonna

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work.

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So what are we gonna do here? There's another rule called the quotient rule.

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let me just tell you what it is we have a quotient of two functions turns out

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the derivative. The numerator is exactly like the product rule, f prime times second

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function,

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but it's really the denominator function, except we have a minus here instead of a

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plus.

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00:26:18,620 --> 00:26:23,280
And then the denominator turns out to be g of x squared.

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The derivative of a quotient resembles the product rule in the numerator,

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except it uses subtraction instead of addition, and the denominator is squared.

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To prove it, we would start with the definition of the derivative.

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And if you can prove it by the end of the class, we'll have time at the end of the

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class.

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You get extra credit.

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I rarely ever offer extra credit.

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Is it going to be similar to the proof of the product rule?

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It's going to be very similar.

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We would add and subtract a carefully chosen expression.

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But then it's a little bit more complicated, now we got fractions.

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But let's not do it right now.

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We'll have plenty of time at the end of class.

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Let's just use it.

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So f of x, what was it? x cubed over sine of x?

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I

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Often just set up parentheses like that

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You can do it in your head if you want and just go right to the answer

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but it's the derivative of the numerator, the first, I call it the first function,

235
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times the denominator minus

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first function times the derivative of the second,

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just like the product rule but with a subtraction, and then we simply divide by the

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denominator

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squared.

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The derivative of x cubed is 3x squared by the power rule.

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This gives 3x squared sine x minus x cubed cosine x, all over sine squared x.

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That's it.

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Easy peasy.

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Okay.

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00:29:37,020 --> 00:29:38,860
Could we simplify this at all?

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00:29:40,960 --> 00:29:48,110
I mean we could manipulate it, have two different fractions, but if there was like

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an x squared

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00:29:49,540 --> 00:29:55,345
here or something like that, then we could possibly simplify some x's, but we're

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done

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here.

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Okay.

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00:30:18,720 --> 00:30:22,380
We talked about this one last week, something like that.

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And we weren't able to find the derivative

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But how can I rewrite tangent

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Sine over cosine and now we have a quotient rule

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So we're gonna find the derivative by the quotient rule

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The derivative of sine is cosine.

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The derivative of cosine is negative sine.

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Can we simplify this at all?

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How can we simplify this?

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Can I do this?

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Definitely not!

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You go to math jail, you do that.

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You can't simplify one term with a factor.

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It has to be a common factor.

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But this simplifies to what?

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Sine squared plus cosine squared equals one.

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The Pythagorean identity.

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That is the derivative.

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However, what is one over cosine?

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Is secant.

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And so we normally write it as secant squared.

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We do want to just memorize this from now on.

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The derivative of tangent is secant squared. Now we know the derivatives of sine,

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cosine, and tangent.

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Let's do one more and we'll start our homework.

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Let's find the derivative of secant x.

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How am I going to start this?

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Rewrite secant x as one over cosine x.

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Change it to 1 over cosine.

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And what rule are we going to use?

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Quotient rule.

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Quotient rule.

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Go ahead and see if you can do this one on your own.

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Zero times cosine x minus one times negative sine x, over cosine squared x, gives

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sine x over cosine squared x.

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That is the derivative of secant x.

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However, we normally don't write it like that.

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we normally change this to...

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to this...

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and then what is one over cosine?

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that is secant...

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And sine over cosine is tangent

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Okay, that's how we normally write it that's the one we just memorize from now on

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Derivative of secant

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There's two more we're gonna do next week.

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Cotangent and cosecant are next; then we will have all six trigonometric

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derivatives.

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Okay, any questions on this?

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We're gonna stop here.