Chain rule
More examples using multiple rules More examples of taking derivatives
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- Now that you've been introduced into some of the other
- functions that we can take a derivative of, we can now
- apply them using the chain and the product rule.
- So let's do some fun derivatives.
- And I think derivatives is all about exposure, it's
- all about practice.
- So I just encourage you to do as much practice as possible.
- And it's actually in some ways a pretty mechanical thing to do
- and it's easier than a lot of the math that you've
- learned before.
- Just maybe initially looks a little abstract.
- So let's say that f of x is equal to let's say the sin
- of 3x to the fifth plus 2x.
- So what is f prime of x?
- What is the derivative of this function?
- Well, we use the chain rule again.
- We take the derivative of the inside.
- So what's the derivative of the inside?
- Well, that's just 5 times 3 is 15 x to the fourth plus 2.
- And then we take the derivative of the larger function.
- In the last presentation we learned the derivative
- of sin is what?
- It's cosine.
- So it's times the cosine of this expression right here.
- 3x to the fifth plus 2x.
- Pretty painless, no?
- Let's mix it up even more.
- Let's say that-- let me switch colors just to
- not be monotonous.
- I'll pick powder blue.
- Very nice.
- Let's say that y-- and I'm going to switch notation on
- purpose that you get used to the various notations
- you can use.
- Let's say that y is equal to-- let me think of something
- good-- e to the x times cosine to the fifth of x.
- That looks daunting to me.
- Let's see if we can break it down using the product
- and the chain rules.
- We want to figure out dy/dx.
- We want to figure out the rate at which y
- changes relative to x.
- Or the derivative.
- Find the derivative of both sides.
- Well, let's use the product rule.
- Well, we're going to have to use the chain and
- the product rules.
- So first we take the derivative of this first term, and once
- again we learned in the last presentation the most amazing
- fact, one of the most amazing facts in the universe that the
- derivative of e to the x is what?
- It is e to the x.
- Blows my mind.
- e to the x.
- Once again I've taken the derivative, and it's
- the same expression.
- Amazing.
- And then I multiply it times a second expression.
- Cosine to the fifth of x.
- And now to that I add the derivative of the
- second expression.
- Now this will be a little bit more interesting.
- So this is cosine of x to the fifth.
- This is just another way of writing cosine of
- x to the fifth power.
- And I think that'll make it a little bit more clear, that
- this cosine superscript 5, this is really just
- cosine to the fifth x.
- This means cosine of x to the fifth.
- So now the derivative is a little bit clearer.
- We can use the chain rule-- and once again, we're just
- working on this right half.
- We take the derivative of the inside.
- What's the derivative of cosine of x?
- Yep you're right.
- Well, I don't know, I didn't hear you so I don't know.
- I'll assume you're right.
- The derivative of cosine of x is minus sin of x, and that's
- something you should memorize, although you should prove
- it to yourself as well.
- So we take the derivative of the inside minus sin.
- Derivative of cosine of x is minus sin of x, and then we
- take the derivative of the outside.
- We're just doing the chain rule.
- So it's 5 cosine to the fourth of x.
- So there we took the derivative of this piece, and then we have
- to multiply times this first piece.
- So that times e to the x.
- Interesting.
- You can simplify this if you want, but you get the point.
- I mean simplifying it from this point is really
- just kind of algebra.
- And I think you get the idea.
- Actually everything we're doing is algebra.
- If you realize it looks like something fairly complicated,
- but we just use the chain and the product rules.
- Let's do some more.
- I will now switch to magenta.
- We want to take the derivative dy/dx of-- let's see,
- some big expression.
- let me do something creative.
- Let's say the natural log of x over 3x plus 10.
- So the natural log of x over 3x plus 10.
- So you could use the quotient rule if you took the time to
- memorize it, which I've never taught you because it's really
- just the product rule.
- So I like to just rewrite this as the product rule.
- So they're the same thing.
- Once again we're taking the derivative, so I'm not going to
- keep rewriting this, but this is the same thing as taking the
- derivative of the natural log of x times 3x plus ten to
- the negative 1 power.
- 3x plus 10 in the denominator is the same thing as 1 over 3x
- plus 10, which is the same thing as 3x plus 10 to
- the negative 1 power.
- Now we can use the combination of the product and the
- chain rules, and we can solve this sucker.
- So let's do it.
- So we take the derivative of this first term the natural log
- of x-- and we learned in the last presentation the
- derivative of the natural log of x is 1/x, which is
- pretty cool in of itself.
- And we multiply that times a second term.
- So time 3x plus 10 to the negative 1 power.
- And to that we add the derivative of the second term,
- and we're going to multiply that times the first term.
- So first we're going to have to use the chain rule.
- We take the derivative of the inside.
- Well the derivative of the inside's easy.
- The derivative of 3x x plus 10.
- That's just 3.
- And then we take the derivative of the whole thing,
- so it's negative 1.
- That's 3 times negative 1 times that whole
- expression to the minus 2.
- 3x plus 10.
- And of course this whole thing times the natural log of x.
- We could simplify that.
- Let's see.
- This is 1/x and 3x plus 10 to negative 1.
- So we could rewrite this as 1 over x 3x plus 10.
- Let's see.
- Plus 3 times minus 1.
- So we could say that's the same thing as minus 3 ln of
- x over 3x plus 10 squared.
- I think you see how I got from here to here.
- I just manipulated the exponents and multiplied some
- numbers, et cetera, et cetera.
- Let's do one more.
- Just hit the point home.
- You really have the tools now at your disposal to do a
- lot of derivative problems.
- Probably most of the derivative problems you'll face in the
- first 1/2 year of calculus.
- I'm going to switch to green.
- Let's say y-- actually I'm tired of y.
- Let's say that p is equal to-- I don't know.
- Sin of x over cosine of x.
- Let's figure out what dp/dx.
- The rate at which p changes to x.
- So once again this is the same thing as sin of x times cosine
- of x to the negative 1.
- So we can just do the product and chain rules.
- So the derivative of the first term.
- Derivative of sin of x is cosine of x, times
- the second term.
- Times cosine of x to the minus 1.
- And then to that we add the derivative of the second term.
- We have to use the chain real here.
- So we take the derivative of the inside.
- Well derivative of cosine x is minus sin of x.
- And then times the derivative of the outside.
- well that's minus 1 cosine of x to the minus 2, and then we
- multiply that times the first term, sin of x.
- So let's simplify that.
- So this is cosine of x divided by cosine of x.
- You see how this cancels out?
- This is cosine of x over cosine of x, so this is equal to 1.
- Cosine of x divided cosine of x is 1.
- And then this minus sin cancels out with this minus sin, and
- we have sin times sin over cosine squared.
- So this is equal to 1.
- I'm going kind of fast because I'm about to run out of time,
- but I think you get what I'm doing.
- So this is sin squared x over cosine squared x, which
- is actually equal to-- what's sin over cosine?
- 1 plus tan squared x.
- And if you know your trig identities, that equals 1
- over cosine squared of x.
- And, of course, what we just proved is that the derivative
- of the tangent of x is equal to the secant squared of x.
- I hope I thoroughly confused you in that last problem.
- I'll see you in the next presentation.
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At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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