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Now that you've been introduced into some of the other

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functions that we can take a derivative of, we can now

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apply them using the chain and the product rule.

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So let's do some fun derivatives.

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And I think derivatives is all about exposure, it's

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all about practice.

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So I just encourage you to do as much practice as possible.

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And it's actually in some ways a pretty mechanical thing to do

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and it's easier than a lot of the math that you've

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learned before.

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Just maybe initially looks a little abstract.

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So let's say that f of x is equal to let's say the sin

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of 3x to the fifth plus 2x.

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So what is f prime of x?

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What is the derivative of this function?

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Well, we use the chain rule again.

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We take the derivative of the inside.

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So what's the derivative of the inside?

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Well, that's just 5 times 3 is 15 x to the fourth plus 2.

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And then we take the derivative of the larger function.

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In the last presentation we learned the derivative

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of sin is what?

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It's cosine.

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So it's times the cosine of this expression right here.

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3x to the fifth plus 2x.

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Pretty painless, no?

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Let's mix it up even more.

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Let's say that-- let me switch colors just to

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not be monotonous.

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I'll pick powder blue.

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Very nice.

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Let's say that y-- and I'm going to switch notation on

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purpose that you get used to the various notations

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you can use.

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Let's say that y is equal to-- let me think of something

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good-- e to the x times cosine to the fifth of x.

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That looks daunting to me.

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Let's see if we can break it down using the product

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and the chain rules.

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We want to figure out dy/dx.

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We want to figure out the rate at which y

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changes relative to x.

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Or the derivative.

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Find the derivative of both sides.

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Well, let's use the product rule.

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Well, we're going to have to use the chain and

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the product rules.

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So first we take the derivative of this first term, and once

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again we learned in the last presentation the most amazing

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fact, one of the most amazing facts in the universe that the

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derivative of either the x is what?

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It is e to the x.

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Blows my mind.

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e to the x.

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Once again I've taken the derivative, and it's

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the same expression.

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Amazing.

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And then I multiply it times a second expression.

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Cosine to the fifth of x.

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And now to that I add the derivative of the

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second expression.

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Now this will be a little bit more interesting.

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So this is cosine of x to the fifth.

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This is just another way of writing cosine of

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x to the fifth power.

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And I think that'll make it a little bit more clear, that

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this cosine superscript 5, this is really just

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cosine to the fifth x.

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This means cosine of x to the fifth.

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So now the derivative is a little bit clearer.

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We can use the chain rule-- and once again, we're just

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working on this right half.

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We take the derivative of the inside.

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What's the derivative of cosine of x?

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Yep you're right.

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Well, I don't know, I didn't hear you so I don't know.

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I'll assume you're right.

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The derivative of cosine of x is minus sin of x, and that's

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something you should memorize, although you should prove

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it to yourself as well.

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So we take the derivative of the inside minus sin.

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Derivative of cosine of x is minus sin of x, and then we

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take the derivative of the outside.

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We're just doing the chain rule.

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So it's 5 cosine to the fourth of x.

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So there we took the derivative of this piece, and then we have

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to multiply times this first piece.

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So that times e to the x.

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Interesting.

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You can simplify this if you want, but you get the point.

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I mean simplifying it from this point is really

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just kind of algebra.

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And I think you get the idea.

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Actually everything we're doing is algebra.

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If you realize it looks like something fairly complicated,

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but we just use the chain and the product rules.

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Let's do some more.

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I will now switch to magenta.

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We want to take the derivative dy/dx of-- let's see,

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some big expression.

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let me do something creative.

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Let's say the natural log of x over 3x plus 10.

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So the natural log of x over 3x plus 10.

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So you could use the quotient rule if you took the time to

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memorize it, which I've never taught you because it's really

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just the product rule.

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So I like to just rewrite this as the product rule.

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So they're the same thing.

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Once again we're taking the derivative, so I'm not going to

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keep rewriting this, but this is the same thing as taking the

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derivative of the natural log of x times 3x plus ten to

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the negative 1 power.

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3x plus 10 in the denominator is the same thing as 1 over 3x

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plus 10, which is the same thing as 3x plus 10 to

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the negative 1 power.

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Now we can use the combination of the product and the

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chain rules, and we can solve this sucker.

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So let's do it.

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So we take the derivative of this first term the natural log

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of x-- and we learned in the last presentation the

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derivative of the natural log of x is 1/x, which is

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pretty cool in of itself.

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And we multiply that times a second term.

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So time 3x plus 10 to the negative 1 power.

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And to that we add the derivative of the second term,

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and we're going to multiply that times the first term.

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So first we're going to have to use the chain rule.

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We take the derivative of the inside.

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Well the derivative of the inside's easy.

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The derivative of 3x x plus 10.

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That's just 3.

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And then we take the derivative of the whole thing,

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so it's negative 1.

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That's 3 times negative 1 times that whole

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expression to the minus 2.

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3x plus 10.

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And of course this whole thing times the natural log of x.

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We could simplify that.

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Let's see.

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This is 1/x and 3x plus 10 to negative 1.

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So we could rewrite this as 1 over x 3x plus 10.

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Let's see.

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Plus 3 times minus 1.

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So we could say that's the same thing as minus 3 ln of

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x over 3x plus 10 squared.

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I think you see how I got from here to here.

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I just manipulated the exponents and multiplied some

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numbers, et cetera, et cetera.

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Let's do one more.

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Just hit the point home.

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You really have the tools now at your disposal to do a

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lot of derivative problems.

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Probably most of the derivative problems you'll face in the

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first 1/2 year of calculus.

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I'm going to switch to green.

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Let's say y-- actually I'm tired of y.

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Let's say that p is equal to-- I don't know.

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Sin of x over cosine of x.

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Let's figure out what dp/dx.

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The rate at which p changes to x.

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So once again this is the same thing as sin of x times cosine

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of x to the negative 1.

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So we can just do the product and chain rules.

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So the derivative of the first term.

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Derivative of sin of x is cosine of x, times

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the second term.

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Times cosine of x to the minus 1.

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And then to that we add the derivative of the second term.

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We have to use the chain real here.

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So we take the derivative of the inside.

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Well derivative of cosine x is minus sin of x.

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And then times the derivative of the outside.

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well that's minus 1 cosine of x to the minus 2, and then we

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multiply that times the first term, sin of x.

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So let's simplify that.

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So this is cosine of x divided by cosine of x.

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You see how this cancels out?

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This is cosine of x over cosine of x, so this is equal to 1.

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Cosine of x divided cosine of x is 1.

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And then this minus sin cancels out with this minus sin, and

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we have sin times sin over cosine squared.

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So this is equal to 1.

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I'm going kind of fast because I'm about to run out of time,

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but I think you get what I'm doing.

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So this is sin squared x over cosine squared x, which

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is actually equal to-- what's sin over cosine?

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1 plus 10 squared x.

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And if you know your trig identities, that equals 1

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over cosine squared of x.

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And, of course, what we just proved is that the derivative

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of the tangent of x is equal to the secant squared of x.

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I hope I thoroughly confused you in that last problem.

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I'll see you in the next presentation.

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