I've been requested to do some more implicit differentiation
or differentiation of implicit functions by I believe his name
is [? Akash ?], so I will do some more.
And these are actually the problems that he sent to me and
they looked good, so those are the ones I'll do.
The first problem he gave me, and I'll do it in green, says y
to the third power minus xy squared plus cosine
xy is equal to 2.
So like all differentiation of implicit functions-- and of
course in these examples I'm going to-- except for the last
one-- I'm going to find the derivative of y
with respect to x.
So in all of these I want to figure out what
dy/dx is equal to.
Well to do that just like in the previous video we just take
the derivative with respect to x on both sides of
this equation.
So we do the derivative operator I guess
we could call it.
So we're going to take it on this side.
ddx.
And we're going to take the derivative of everything.
Then we're going to take the derivative on the right hand
side, the derivative of 2, which we know hopefully
by this point is 0.
And I'll do it a little bit quicker than in last video, but
anyway send me a note if you think I did it too fast.
So let's take the derivative of each of these terms.
So the derivative with respect to x of y cubed.
Well it's the derivative with respect to y, we're just
doing the chain rule.
So it's 3y-- that's a y, not a g-- 3y squared.
So the derivative of this function with respect to y
times the derivative of y with respect to x.
dy/dx.
That's just the chain rule.
And then we have minus the derivative of this
with respect to x.
Well this is a little interesting, because here we
have x times y squared, which we're arguing is maybe a
function of what's an implicit function of x.
So how do we do that?
What we'll do-- let me put the subtraction sign outside and
then let's do the derivative inside, and I'll do it
inside some brackets.
So let's do the product rule here.
We have x times y squared.
The product rule says it's the derivative of the first
function times a second function plus the derivative of
the second function times the first function.
So the derivative of the first function with respect to x--
remember what we're taking the derivative with respect to--
well the derivative of x is just 1.
1 times y squared plus the derivative of
the second function.
Well what's the derivative of the second function?
Well it's going to be 2y dy/dx, and that just comes
from the chain rule.
The derivative with respect to y times the derivative
of y with respect to x.
That's the chain rule.
So that's the derivative of y squared, and then we multiply
that times the first function, which is x.
Times x.
Hopefully that didn't confuse you.
Just took the derivative of x times y squared.
This is the product rule.
Plus the derivative of y squared, which
was this times x.
OK now we're ready to move on.
OK.
Now this is interesting.
This is interesting.
We have to take the derivative of cosine of xy.
Well once again we just do the chain rule.
And maybe I'll actually break it up into a couple of steps
here just so that you don't get completely confused.
So what does the chain rule say if we have you know f of g
of x of whatever whatever?
We essentially have to take the derivative of the inside
function and multiply-- or well I guess you could
do it two ways.
You could say we take the entire derivative with respect
to whatever this function is.
You know if g was equal to xy we could say the derivative of
cosine of g with respect to g.
And then we would multiply that times the derivative
of this with respect to x.
If that confuses you, I think it'll make more sense when
I actually do the example.
Actually let me write out like just the way I said it, because
I think that could maybe even clear up some confusion
over the chain rule.
Or maybe it'll confuse you further.
So what are we doing?
We're going to do the derivative-- let me do this
in a different color.
So let's take derivative of the outside function
with respect to xy.
So what is that?
What's the derivative of cosine of x?
Or let's say if it was g is equal to xy.
If g is equal to xy, then that expression there
is cosine of g.
So what's the derivative of this with respect to g?
Sorry, dg.
Well that's just minus sin of g, right?
So let me do that in a different color.
So it's minus sin, and g was just xy.
So that's the derivative of you can kind of say
of the outside function.
And then we go to the inside, and we have to multiply this
times the derivative of this inside function.
Let me just write that like this.
Times the derivative with respect to x--that's what
we're doing-- of xy.
And I'm going to leave it like that for now because
that's going to take a little bit of space.
And then that equals the derivative of 2
with respect to x.
Well 2 never changes, right?
It's a constant, so it's derivative is 0.
Let's see if we can simplify this a little bit, and
we'll come back to that, let's just write this.
So we have 3g-- no I'm confusing myself with
my bad penmanship.
3y-- all that talk of g got me confused-- dy/dx.
3y squared dy/dx minus y squared minus-- I'm just
distributing the minus-- xy minus 2xy dy/dx
minus sin of xy.
And then this whole thing times the derivative of
xy with respect to x.
Well we can just do the product rule here.
What's the derivative of xy with respect to x?
Well the derivative of x is 1, so it's 1 times y.
Derivative of the first times the second function.
Plus the derivative of the second function
with respect to x.
So what's the derivative of y with respect to x?
Well it's just dy/dx, right?
dy/dx times the first function.
That's just the product rule.
Review of the product rule if that was a little confusing.
It's equal to 0.
And now we can attempt to simplify.
So let's see if I can do it.
So we have 3y squared dy/dx minus y squared
minus 2xy dy/dx.
Now this one's interesting, so let me distribute
the sin of xy.
This is just a 1, so it's minus y sin of xy minus x dy/dx sin
of xy-- I'm just simplifying, this is just algebra.
So y is equal to 0.
Now let's group all of the terms that have a dy/dx in them
and distribute out the dy/dx.
So this is equal to dy/dx times this 3y squared minus two xy.
I'm just undistributing where I'm dividing out the dy/dx.
So let me actually show you the terms.
So this has a dy/dx, this has a dy/dx.
And this is just algebra.
This has a dy/dx, so let me separate those out.
So it's 3y squared minus 2xy minus x sin of xy.
Well that's all of the ones with the dy/dx on them.
And then these terms-- and I'm just skipping a couple of
steps because this is algebra right now.
The non dy/dx terms-- that term, and I guess there's only
two of them-- I'm just going to add to the other
side of equation.
So I'm just going to bring them over to the other side.
Hopefully that step doesn't confuse you.
This is just algebraic simplification at this point.
I could have written minus y squared minus y sin xy and then
added them on the other side, but I'm about to run out of
space and time, so I'm just going to add these to both
sides of the equation.
So that equals y squared plus y sin of xy.
I think I've covered all of the terms.
And now since our goal is to find the derivative of y with
respect to x, we can divide both sides of this equation by
this term right here, and our final answer is-- let me
do it in a good color.
dy/dx.
The derivative of y with respect to x is equal to that
y squared plus y sin of xy divided by this-- 3y squared
minus 2xy minus x sin of xy.
And there you have it.
That was a good implicit differentiation problem and if
you understood how to do that, you understand how to do
implicit differentiation.
And I hope I didn't make any careless mistakes, and it's a
good exercise for you to try to do it again on your own.
See you soon.