I've been asked implicitly differentiate the equation
tangent of x over y is equal to x plus y.
And I've done several implicit differentiation videos, but
this tends to be one of the biggest sources of pain for
first year calculus students.
So I thought I would give at least another example.
It never hurts to see as many as possible.
So let's do this one.
So to implicitly differentiate this, we just apply the
derivative with respect to x operator to both sides
of the equation.
The derivative with this respect to x -- the derivative
of the left side with respect to x is the same as the
derivative of the right side with respect to x.
The right side's going to be very straightforward, but
the left side is a little bit tricky.
So let's do that on the side over here.
Let me write the left hand side a little bit differently.
I'm going to do it in a different color.
Let me say that a is equal to the tangent of b.
And let me say that b is equal to x over y.
Then a is clearly the same thing.
I mean if I just substituted b back in here, a, this
whole thing I could re-write as just a.
So if we're taking the derivative of a with respect
to x, that's what we want to do right here.
Let me just take the derivative of both sides of this.
This would be the derivative of a with respect to x is equal to
the derivative of x with respect to x.
Well that's pretty straightforward, that's just 1.
Plus the derivative of y with respect to x.
So let me write it like this.
I'll write the derivative operator, the derivative
oh y with respect to x.
That's all we did.
We just applied the derivative operator to y, and we don't
know what this thing is, we're going to solve for it.
But obviously, I can't just leave this here, the derivative
of a with respect to x.
We just solved for a, and a is just this thing
right here, right?
a is tangent of b, and b is just y over x.
The reason why I wrote it this way is because I wanted to show
you that when you take the derivative of this, it just
comes out of the chain rule.
It's not some type of new voodoo magic that you
haven't learned yet.
So the derivative -- let me just write down the
chain rule right here.
The derivative of a with respect to x is equal to the
derivative of a with respect to b times the derivative
of b with respect to x.
That's just the chain rule and it's very easy to remember,
because the db's cancel out and you're just left with the
derivative of a with respect to x, if you just treated these
like regular fractions.
So what's the derivative of a with respect to b?
Well, that's just 1 over cosine squared of b.
And if you don't have that memorized, it's actually not
too hard to prove to yourself if you just write this as sine
of b over cosine of b, but this tends to be one of the trig
derivatives that most people memorize.
I think I've already made a video where I proved this.
And some books still write this as secant squared of b, but we
know that secant squared is the same thing as 1 over
cosine squared.
I like to keep it in kind of the fundamental trig functions,
or trig ratios as opposed to things like secant
and cosecant.
Then what's the derivative of b with respect to x?
So this is pretty interesting.
Let me re-write b, actually.
Let me write b is equal to x times y to the minus 1.
So the derivative of b with respect to x, we could do a
little bit of chain rule right here.
We could say -- let me write this -- the derivative of b
with respect to x is equal to the derivative of x times
y to the negative 1.
So the derivative of x is 1.
times y to the negative 1 plus the derivative of y -- so
let me just write this.
Plus the derivative with respect to x of y to
the minus 1 times the first term, times x.
So this thing right here, and clearly I haven't completely
simplified it yet.
I still have to figure out what this thing is here.
But I just simply applied the product rule here.
Your derivative of the first term, derivative of x is 1
times the second term plus the derivative of the second
term times the first term.
That's all I did there.
So the derivative of b with respect to x is just
this thing right there.
So it equals -- let me do it in the yellow -- so it's times --
oh, I'll do it in the blue since I already wrote it.
This is the blue, derivative of b with respect to x is y to the
minus 1, or 1 over y plus the derivative with respect to
x of 1 over y times x.
So let me write that down here.
So we just figured out, or we're almost done figuring
out, what the derivative of a with respect to x is, and we
could throw that in there.
But we're not done.
What's the derivative of 1 over y with respect to x?
Well, do the chain rule again.
And I want to be very clear with this.
I know this might seem a little bit cumbersome what I'm doing
here, but I think it might make a little bit of sense.
Let me just set c is equal to 1 over y.
So the derivative of c with respect to x, just from the
chain rule, is equal to the derivative of c with respect
to y times the derivative of y with respect to x.
What's the derivative of c with respect to y?
Well this is the same thing as -- I could re-write
this as y to the minus 1.
So it's minus y to the minus 2 power.
That's what this thing is.
This thing is that right there.
And I don't know what the derivative of y with
respect to x is.
That's what we're trying to solve for.
So it's that times the derivative of y
with respect to x.
That just comes out of the chain rule.
So this thing right here, this is the derivative of this thing
with respect to x, which is the same thing as derivative
of c with respect to x.
So I can write this little piece right here, I can
re-write this little piece as minus y to the minus 2 dy
dx, and then, of course, that there is times x.
And then we had the plus 1 over y, and all of that was times
the 1 over cosine squared of b.
So now we've simplified this a good bit.
I hope going into the chain rule didn't confuse you,
because I really want to hit the point home that all of
these implicit differentiation problems, these dy dx's just
don't, it's not some rule that you should memorize.
They come out naturally from the chain rule.
So we solved da dx, that is equal to this
expression right here.
Let me write it, it's equal to 1 over cosine squared of b.
Well what's b?
I wrote it's cos x over y.
Cosine squared of x over y times all of this stuff over
here, times all of this mess.
1 over y plus, or maybe I should say minus, minus -- if I
just simplify this, this is x over y squared times dy dx.
Then that is equal to the right hand side.
It is equal to 1 plus dy dx.
And now all we have to do is solve for dy dx.
So let me just review how we got here.
I went through the chain rule at every step of the way, but
once you get the hang of it, you can literally just go
straight down this way.
The way you think about it is -- the right hand
side I think you get it.
The derivative of x is 1, the derivative of y with respect
to x, well that's just dy dx.
But the left hand side, you take the derivative of
the whole thing with respect to x over y.
So that's just the derivative of tangent is 1 over
cosine squared.
So it's 1 over cosine squared of x over y, and you multiply
that times the derivative of x over y with respect to x.
And the derivative of x over y with respect to x is the
derivative of-- and it gets complicated, that's why it's
good to do it on the side here -- but it's the derivative of
x, which is 1 times 1 over y.
Which is that term plus the derivative of 1 over Y with
respect to X, which is minus 1 over y squared dy dx, from
the chain rule, times dx.
That's why it was good to do over to the side so we don't
make a careless mistake.
But once you get used to it you could actually do that in your
head, and of course, that equals the right hand side.
So from here on out it's just pure algebra.
Just to solve for our dy dx.
So a good place to start is to multiply both sides of this
equation times cosine squared of x over y.
So obviously, that'll turn to 1 on this side.
And the left hand side will be 1 over y minus x over y squared
dy dx is equal to -- I have to multiply both side of the
equation times this denominator right here -- is equal to
cosine squared of x over y plus cosine squared of
x over y dy dx.
Now what can we do.
We can subtract this cosine squared of x over y from both
sides of the equation, and we get 1 over y minus cosine
squared of x over y.
All I did is I subtracted this from both sides of the
equation, so essentially I moved it over to the
left hand side.
What I'm trying to do is I'm going to try to separate the
non dy dx terms from the dy dx terms.
So I want to bring this dy dx term over to
the right hand side.
So let me add x over y squared dy dx to both sides.
So then that is equal to x over y -- let me write that in the
color that I originally wrote it in, a slightly
different color.
So that is x over y squared -- I'll write the dy dx in orange.
dy dx, and then you have this term, plus cosine squared
of x over y dy dx.
I think we're in the home stretch.
Let's factor this dy dx out from the right hand side.
So this is equal to dy dx times x over y squared plus
cosine squared of x over y.
And that is equal to this thing over here, it's equal to
1 over y minus cosine squared of x over y.
Now to solve for dy dx, we just have to divide both sides of
this equation by this expression right here.
And then what do we get?
We get, if we just divide both sides by that, we get 1 over y
minus cosine squared of x over y divided by this whole
business right there.
x over y squared plus cosine squared of x over y is
equal to our dy dx.
And then we're done.
We just applied the chain rule multiple times and we were able
to implicitly differentiate tangent of y over x is
equal to x plus y.
The hard part really is getting to this step right here.
After this step it's literally just pure algebra just to solve
for the dy dx's, and then you get that answer right there.
Anyway, hopefully you found that useful.