Trig Implicit Differentiation Example

Implicit differentiation example that involves the tangent function

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.