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Welcome back.

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Let's do some more derivative problems.

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Let's say I want to figure out the derivative d over dx of--

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and let me give something that looks a little bit different--

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x to the third minus 5x to the fifth, all of that to the third

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power over 2x plus 5 to the fifth power.

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This is a parentheses.

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This is just saying that I want to take the derivative of

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this entire expression.

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So you're saying Sal, we've never learn how to do this,

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you have something in the numerator, you have something

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in the denominator I don't know what to do next.

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Well let's just rewrite this.

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Actually in your calculus textbooks there's something

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called the quotient rule, which I think is mildly lame, because

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the quotient rule is just the product rule where you have a

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negative exponent and they make it another rule, and they

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clutter your brain.

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So instead of using the quotient rule, we're just

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going to rewrite this bottom expression as a product,

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and then we can use the product rule.

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So this is the same thing as taking the derivative of x to

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the third minus 5x to the fifth, all of that to the third

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power, times 2x plus 5 to what?

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The minus fifth power.

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And now we can use the product rule.

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Take the derivative of the first term-- and the derivative

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of the first term isn't a joke-- you take the derivative

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of the inside first, let's do the chain rule, derivative

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of the inside first.

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That is 3x squared minus 25x to the fourth times the derivative

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of the outside, 3 times this entire expression x to the

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third minus 5x to the fifth.

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And then all of that, take this exponent down one to the

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squared, and then multiply it times this whole term.

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So 2x plus 5 to the minus fifth.

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And then to that we add the derivative of

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this term, so plus.

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So the derivative of this term we take the derivative of the

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inside, which is pretty easy.

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It's just 2 times the derivative of the outside,

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which is minus 5.

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And just so you know I didn't skip a step, the derivative of

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2x plus 5, the derivative of 2x is 2, derivative of 5 is 0.

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So the derivative of 2x plus 5 is just 2.

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So it's 2 times minus 5 2x plus 5.

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We just keep that the same to the minus fifth power, and then

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we multiply it times this first expression, x to the third

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minus 5x to the fifth to the third power.

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I know that's really messy and you'll probably not see

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problems this messy, but I just wanted to show you that the

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product rule we learned-- it's actually the product and the

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chain rule-- they can apply to a lot of different problems,

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and even though you hadn't seen something like this where you

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had numerator and a denominator, you can easily

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rewrite what you had in the denominator as a

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negative exponent.

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And then of course it's just the product for when you don't

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have to memorize that silly thing called the quotient rule.

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So with that out of the way, I'm now going to introduce you

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to some common derivatives of other functions.

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And these things are actually normally included in the inside

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cover of your calculus book, and they're just good to

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know, good things to know.

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And maybe in a later presentation I'll actually

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prove these things.

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You should never take things at face value.

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So you should to some degree memorize these, although you

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should prove it to yourself first.

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So the derivative of e to the x-- and I find this to be

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amazing. e shows up all sorts of crazy places in mathematics,

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and it's you know the strange number 2.7 whatever, whatever

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and it has all sorts of strange properties.

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And I think this is one of the most bizarre properties of e.

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The derivative of e to the x.

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So if I want to figure out the slope of any point along

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the curve e to the x-- this just might blow your mind.

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I think the more you think about it, the more it'll blow

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your mind-- is e to the x.

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That's amazing.

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At any point along the curve e to the x, the slope of

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that point is e to the x.

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Just to hit the point home.

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I'm diverging, a little bit.

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But if I said f of x is equal to e to the x, right?

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And let's say f of 2 is equal to e squared.

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And I asked you, friend-- I don't know your name-- what

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is the slope of e to the x at the point 2,e squared.

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And you could say Sal, the slope at that

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point is e squared.

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That blows my mind that it's a function where the slope at

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any point on that line is equal to the function.

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And it's e. e shows up all sorts of places.

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I might do a whole series of presentations called the

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magic of e, because e shows up all over the place.

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Well I don't want to diverge too much, so that's

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pretty amazing.

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Next I'm going to show you what I think is probably the second

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most amazing derivative-- and I don't think this has been fully

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explored in mathematics yet, because this also blows my

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mind-- is that the derivative of the natural log of x, right.

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So the natural log is just the logarithm with base e, and I

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hope you remember your logarithms.

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So what's the derivative of the natural log of x?

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So once again this is e related.

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Well it's 1/x.

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That also blows my mind.

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Because think about it.

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Let's draw a bunch of functions.

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If I said the derivative of x to the minus 3 is

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minus 3x to the minus 4.

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The derivative of x to the minus 2 is minus

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2x to the minus 3.

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The derivative of x to the minus 1 is minus

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1 x to the minus 2.

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The derivative of x to the 0-- well this is just 1, right?

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The derivative of x to the 0 is just 1, so the derivative is 0.

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The derivative of x is 1, derivative of x squared

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is 2x and so on, right?

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So it's interesting.

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We have this pattern from all the derivatives of all of the

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of kind of the exponents in increasing order where you go

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from x to the minus 4 x to the minus 3, x to the minus 2 and

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then there's no x to the minus 1 here.

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We go straight to x to the 0.

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What happened to x the minus 1?

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What happened to this?

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What function's derivative is x to the minus 1?

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This is bizarre to me.

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Where did it go?

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And it turns out that it's a natural log.

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This I still think about before I go to bed sometimes because

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it is kind of mind blowing.

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And later in another presentation I might

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actually prove this to you.

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But just to know that this is true, that the derivative of

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the natural log of x is 1/x I think is mind blowing.

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And so for now you can just memorize it.

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But both of these are mind blowing.

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The derivative of e to the x is e to the x, and the derivative

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of the natural log of x is 1/x.

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And I'll just do a couple of more just to present them to

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you, and then in the next presentation we'll actually use

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them using the product rule and the chain rule and et

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cetera, et cetera.

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And you might want to rewatch this and memorize them.

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I want to clear image.

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OK.

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And now I'll just do the basic trig functions, and you should

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memorize these as well.

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The derivative of sin of x-- this is pretty easy to

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remember-- is cosine of x.

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So the slope at any point along the [? line ?]

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sin of x is actually the cosine of that point.

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That's also interesting.

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One day I'm going to do this holographically because I

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think that might not be sinking in properly.

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The derivative of cosine of x is minus sin of x.

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There are good to memorize though, because you'll be

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able to recall is quickly on a test and then use it.

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And then finally the derivative of tan of x is equal to 1 over

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cosine square of x which you could also write as the

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secant squared of x.

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You might want to memorize these now, and actually I

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encourage you to explore these things, I encourage you to

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graph each of these functions.

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Graph a function, graph its derivative and look at them,

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and really intuitively understand why the derivative

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function actually does describe the slope of

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the original function.

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And actually I'll probably do a presentation on that.

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But I'm almost out of time in this presentation,

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so just memorize these.

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And memorize the derivative of e to the x, e to the x, and

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the natural log of x is 1/x.

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And in the next presentation we're going to start mixing and

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matching all of these functions, and we can use the

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product and chain rule on them to solve kind of arbitrarily

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complex derivatives.

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Between what we've just seen, we could probably solve 95% of

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the derivative problems you'll see on say the

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calculus AP test.

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I'll see you in the next presentation.

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