0:00:00.000,0:00:00.430


0:00:00.430,0:00:04.050
I've been asked implicitly[br]differentiate the equation

0:00:04.050,0:00:10.390
tangent of x over y is[br]equal to x plus y.

0:00:10.390,0:00:14.150
And I've done several implicit[br]differentiation videos, but

0:00:14.150,0:00:17.440
this tends to be one of the[br]biggest sources of pain for

0:00:17.440,0:00:18.720
first year calculus students.

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So I thought I would give[br]at least another example.

0:00:21.040,0:00:22.860
It never hurts to see[br]as many as possible.

0:00:22.860,0:00:24.290
So let's do this one.

0:00:24.290,0:00:26.680
So to implicitly differentiate[br]this, we just apply the

0:00:26.680,0:00:29.363
derivative with respect to x[br]operator to both sides

0:00:29.363,0:00:29.970
of the equation.

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The derivative with this[br]respect to x -- the derivative

0:00:33.290,0:00:35.420
of the left side with respect[br]to x is the same as the

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derivative of the right[br]side with respect to x.

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The right side's going to be[br]very straightforward, but

0:00:42.790,0:00:44.770
the left side is a[br]little bit tricky.

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So let's do that on[br]the side over here.

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Let me write the left hand side[br]a little bit differently.

0:00:52.020,0:00:52.990
I'm going to do it in[br]a different color.

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Let me say that a is equal[br]to the tangent of b.

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And let me say that b[br]is equal to x over y.

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Then a is clearly[br]the same thing.

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I mean if I just substituted[br]b back in here, a, this

0:01:14.860,0:01:18.090
whole thing I could[br]re-write as just a.

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So if we're taking the[br]derivative of a with respect

0:01:20.930,0:01:23.740
to x, that's what we[br]want to do right here.

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Let me just take the derivative[br]of both sides of this.

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This would be the derivative of[br]a with respect to x is equal to

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the derivative of x[br]with respect to x.

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Well that's pretty[br]straightforward, that's just 1.

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Plus the derivative of[br]y with respect to x.

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So let me write it like this.

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I'll write the derivative[br]operator, the derivative

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oh y with respect to x.

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That's all we did.

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We just applied the derivative[br]operator to y, and we don't

0:01:56.520,0:01:58.650
know what this thing is,[br]we're going to solve for it.

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But obviously, I can't just[br]leave this here, the derivative

0:02:01.180,0:02:02.360
of a with respect to x.

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We just solved for a, and[br]a is just this thing

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right here, right?

0:02:05.930,0:02:09.450
a is tangent of b, and[br]b is just y over x.

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The reason why I wrote it this[br]way is because I wanted to show

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you that when you take the[br]derivative of this, it just

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comes out of the chain rule.

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It's not some type of new[br]voodoo magic that you

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haven't learned yet.

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So the derivative -- let[br]me just write down the

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chain rule right here.

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The derivative of a with[br]respect to x is equal to the

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derivative of a with respect[br]to b times the derivative

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of b with respect to x.

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That's just the chain rule and[br]it's very easy to remember,

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because the db's cancel out and[br]you're just left with the

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derivative of a with respect to[br]x, if you just treated these

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like regular fractions.

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So what's the derivative[br]of a with respect to b?

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Well, that's just 1 over[br]cosine squared of b.

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And if you don't have that[br]memorized, it's actually not

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too hard to prove to yourself[br]if you just write this as sine

0:03:07.400,0:03:10.670
of b over cosine of b, but this[br]tends to be one of the trig

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derivatives that most[br]people memorize.

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I think I've already made a[br]video where I proved this.

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And some books still write this[br]as secant squared of b, but we

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know that secant squared is the[br]same thing as 1 over

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cosine squared.

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I like to keep it in kind of[br]the fundamental trig functions,

0:03:25.320,0:03:27.360
or trig ratios as opposed to[br]things like secant

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and cosecant.

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Then what's the derivative[br]of b with respect to x?

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So this is pretty interesting.

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Let me re-write b, actually.

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Let me write b is equal to[br]x times y to the minus 1.

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So the derivative of b with[br]respect to x, we could do a

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little bit of chain[br]rule right here.

0:03:50.470,0:03:53.680
We could say -- let me write[br]this -- the derivative of b

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with respect to x is equal to[br]the derivative of x times

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y to the negative 1.

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So the derivative of x is 1.

0:04:01.300,0:04:07.360
times y to the negative 1 plus[br]the derivative of y -- so

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let me just write this.

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Plus the derivative with[br]respect to x of y to

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the minus 1 times the[br]first term, times x.

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So this thing right here, and[br]clearly I haven't completely

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simplified it yet.

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I still have to figure out[br]what this thing is here.

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But I just simply applied[br]the product rule here.

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Your derivative of the first[br]term, derivative of x is 1

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times the second term plus[br]the derivative of the second

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term times the first term.

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That's all I did there.

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So the derivative of b[br]with respect to x is just

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this thing right there.

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So it equals -- let me do it in[br]the yellow -- so it's times --

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oh, I'll do it in the blue[br]since I already wrote it.

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This is the blue, derivative of[br]b with respect to x is y to the

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minus 1, or 1 over y plus the[br]derivative with respect to

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x of 1 over y times x.

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So let me write that down here.

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So we just figured out, or[br]we're almost done figuring

0:05:04.330,0:05:07.400
out, what the derivative of a[br]with respect to x is, and we

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could throw that in there.

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But we're not done.

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What's the derivative of 1[br]over y with respect to x?

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Well, do the chain rule again.

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And I want to be very[br]clear with this.

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I know this might seem a little[br]bit cumbersome what I'm doing

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here, but I think it might[br]make a little bit of sense.

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Let me just set c is[br]equal to 1 over y.

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So the derivative of c with[br]respect to x, just from the

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chain rule, is equal to the[br]derivative of c with respect

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to y times the derivative[br]of y with respect to x.

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What's the derivative of[br]c with respect to y?

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Well this is the same thing[br]as -- I could re-write

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this as y to the minus 1.

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So it's minus y to[br]the minus 2 power.

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That's what this thing is.

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This thing is that right there.

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And I don't know what the[br]derivative of y with

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respect to x is.

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That's what we're[br]trying to solve for.

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So it's that times[br]the derivative of y

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with respect to x.

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That just comes out[br]of the chain rule.

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So this thing right here, this[br]is the derivative of this thing

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with respect to x, which is the[br]same thing as derivative

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of c with respect to x.

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So I can write this little[br]piece right here, I can

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re-write this little piece as[br]minus y to the minus 2 dy

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dx, and then, of course,[br]that there is times x.

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And then we had the plus 1 over[br]y, and all of that was times

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the 1 over cosine squared of b.

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So now we've simplified[br]this a good bit.

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I hope going into the chain[br]rule didn't confuse you,

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because I really want to hit[br]the point home that all of

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these implicit differentiation[br]problems, these dy dx's just

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don't, it's not some rule[br]that you should memorize.

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They come out naturally[br]from the chain rule.

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So we solved da dx,[br]that is equal to this

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expression right here.

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Let me write it, it's equal to[br]1 over cosine squared of b.

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Well what's b?

0:07:07.880,0:07:10.640
I wrote it's cos x over y.

0:07:10.640,0:07:16.920
Cosine squared of x over y[br]times all of this stuff over

0:07:16.920,0:07:19.840
here, times all of this mess.

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1 over y plus, or maybe I[br]should say minus, minus -- if I

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just simplify this, this is x[br]over y squared times dy dx.

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Then that is equal to[br]the right hand side.

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It is equal to 1 plus dy dx.

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And now all we have to[br]do is solve for dy dx.

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So let me just review[br]how we got here.

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I went through the chain rule[br]at every step of the way, but

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once you get the hang of it,[br]you can literally just go

0:07:58.170,0:07:59.360
straight down this way.

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The way you think about[br]it is -- the right hand

0:08:01.380,0:08:02.033
side I think you get it.

0:08:02.033,0:08:04.380
The derivative of x is 1, the[br]derivative of y with respect

0:08:04.380,0:08:06.560
to x, well that's just dy dx.

0:08:06.560,0:08:09.010
But the left hand side, you[br]take the derivative of

0:08:09.010,0:08:11.630
the whole thing with[br]respect to x over y.

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So that's just the derivative[br]of tangent is 1 over

0:08:14.100,0:08:15.020
cosine squared.

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So it's 1 over cosine squared[br]of x over y, and you multiply

0:08:18.620,0:08:23.530
that times the derivative of[br]x over y with respect to x.

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And the derivative of x over[br]y with respect to x is the

0:08:26.770,0:08:28.970
derivative of-- and it gets[br]complicated, that's why it's

0:08:28.970,0:08:31.590
good to do it on the side here[br]-- but it's the derivative of

0:08:31.590,0:08:34.150
x, which is 1 times 1 over y.

0:08:34.150,0:08:39.680
Which is that term plus the[br]derivative of 1 over Y with

0:08:39.680,0:08:44.200
respect to X, which is minus[br]1 over y squared dy dx, from

0:08:44.200,0:08:46.620
the chain rule, times dx.

0:08:46.620,0:08:48.140
That's why it was good to do[br]over to the side so we don't

0:08:48.140,0:08:49.360
make a careless mistake.

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But once you get used to it you[br]could actually do that in your

0:08:51.410,0:08:53.980
head, and of course, that[br]equals the right hand side.

0:08:53.980,0:08:56.520
So from here on out it's[br]just pure algebra.

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Just to solve for our dy dx.

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So a good place to start is to[br]multiply both sides of this

0:09:01.490,0:09:04.910
equation times cosine[br]squared of x over y.

0:09:04.910,0:09:07.420
So obviously, that'll[br]turn to 1 on this side.

0:09:07.420,0:09:14.970
And the left hand side will be[br]1 over y minus x over y squared

0:09:14.970,0:09:23.690
dy dx is equal to -- I have to[br]multiply both side of the

0:09:23.690,0:09:26.730
equation times this denominator[br]right here -- is equal to

0:09:26.730,0:09:32.530
cosine squared of x over y plus[br]cosine squared of

0:09:32.530,0:09:35.190
x over y dy dx.

0:09:35.190,0:09:39.420


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Now what can we do.

0:09:40.190,0:09:44.210
We can subtract this cosine[br]squared of x over y from both

0:09:44.210,0:09:52.110
sides of the equation, and we[br]get 1 over y minus cosine

0:09:52.110,0:09:53.710
squared of x over y.

0:09:53.710,0:09:55.780
All I did is I subtracted[br]this from both sides of the

0:09:55.780,0:09:57.590
equation, so essentially I[br]moved it over to the

0:09:57.590,0:09:59.040
left hand side.

0:09:59.040,0:10:01.040
What I'm trying to do is I'm[br]going to try to separate the

0:10:01.040,0:10:04.810
non dy dx terms from[br]the dy dx terms.

0:10:04.810,0:10:06.750
So I want to bring this[br]dy dx term over to

0:10:06.750,0:10:07.950
the right hand side.

0:10:07.950,0:10:11.550
So let me add x over y[br]squared dy dx to both sides.

0:10:11.550,0:10:17.260
So then that is equal to x over[br]y -- let me write that in the

0:10:17.260,0:10:21.070
color that I originally wrote[br]it in, a slightly

0:10:21.070,0:10:21.470
different color.

0:10:21.470,0:10:27.110
So that is x over y squared --[br]I'll write the dy dx in orange.

0:10:27.110,0:10:34.120
dy dx, and then you have this[br]term, plus cosine squared

0:10:34.120,0:10:36.880
of x over y dy dx.

0:10:36.880,0:10:40.950


0:10:40.950,0:10:43.000
I think we're in[br]the home stretch.

0:10:43.000,0:10:46.410
Let's factor this dy dx out[br]from the right hand side.

0:10:46.410,0:10:56.770
So this is equal to dy dx[br]times x over y squared plus

0:10:56.770,0:11:01.220
cosine squared of x over y.

0:11:01.220,0:11:04.180
And that is equal to this thing[br]over here, it's equal to

0:11:04.180,0:11:09.250
1 over y minus cosine[br]squared of x over y.

0:11:09.250,0:11:12.240
Now to solve for dy dx, we just[br]have to divide both sides of

0:11:12.240,0:11:15.450
this equation by this[br]expression right here.

0:11:15.450,0:11:16.900
And then what do we get?

0:11:16.900,0:11:21.970
We get, if we just divide both[br]sides by that, we get 1 over y

0:11:21.970,0:11:27.210
minus cosine squared of x over[br]y divided by this whole

0:11:27.210,0:11:28.720
business right there.

0:11:28.720,0:11:36.190
x over y squared plus cosine[br]squared of x over y is

0:11:36.190,0:11:42.150
equal to our dy dx.

0:11:42.150,0:11:43.370
And then we're done.

0:11:43.370,0:11:46.460
We just applied the chain rule[br]multiple times and we were able

0:11:46.460,0:11:50.600
to implicitly differentiate[br]tangent of y over x is

0:11:50.600,0:11:51.600
equal to x plus y.

0:11:51.600,0:11:55.980
The hard part really is getting[br]to this step right here.

0:11:55.980,0:11:59.470
After this step it's literally[br]just pure algebra just to solve

0:11:59.470,0:12:04.730
for the dy dx's, and then you[br]get that answer right there.

0:12:04.730,0:12:07.380
Anyway, hopefully you[br]found that useful.