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 either 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 10 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.