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### Course: Differential Calculus > Unit 3

Lesson 4: Implicit differentiation (advanced examples)# Implicit differentiation (advanced example)

Implicit derivative of (x²+y²)³=5x²y². Created by Sal Khan.

## Want to join the conversation?

- Why, early in the equation solving, was the 5 taken out of the derivative,1:40, and then multiplied back in later,3:56, instead of performing the derivative of 5x^2y^2 when taking the derivative of everything else? Although I understand the the ending result and the concept this small point is a bit confusing to me in relation to the rest of the problem and the prior videos.(20 votes)
- Novamwilliams2,

I'm not sure what to say besides it being a matter of taste. Sal could have chosen to do the product rule using 5x^2 and y^2 as his two functions that were being multiplied. In general I see people factor out any constants and then bring them back after all the calculus is done. I would not be surprised if Sal is not consistent with how he does this.(33 votes)

- How is the derivative of y^2 with respect to x : 2y (dy/dx)?(16 votes)
- It's another chain rule thing, because it applies when you're taking the derivative of something, so y^2 becomes:

(2y^(2-1)) • (derivative of y with respect to x)

or:

2y • (dy/dx)

Similarly, y^1 in the same situation would go through the chain rule, but would cancel itself out via its exponent being zero:

(1y^(1-1)) • (derivative of y with respect to x)

or:

(1) • (dy/dx)(13 votes)

- Why does Sal keep distributing terms like this at3:11? I keep thinking he should be using FOIL, but it seems like he is treating binomials as if they are a single term?(7 votes)
- That's exactly what FOIL does, too: they both use the distributive property.

a(b + c) = ab + ac

(a + b)(c + d) = a(c + d) + b(c + d) = ac + ad + bc + bd

(a + b + c)(d + e) = (a + b + c)d + (a + b + c)e = ad + bd + cd + ae + be + ce

etc.(14 votes)

- He could simplify the end expression abit right? atleast divide it by 2.

Is there a spesific reason he didnt do that?(5 votes)- In calculus, unlike some earlier levels of math, it is not at all uncommon to leave an expression in something other than its simplest form. In fact, even what counts as "the simplest form" may be ambiguous.

So, in Calculus, you think more in terms of whether the form is accurate, informative, useful, or practical. As such, not fully simplifying the answer is not at all unusual.(12 votes)

- For the final answer... if you subbed in the x and y values for a given point to find the slope of the tangent line, wouldn't you get more than one answer? This isn't a function it doesn't pass the vertical line test. There are multiple values for y when given an x value.(6 votes)
- That is exactly why EXPLICIT differentiation doesn't work and you need to use IMPLICIT differentiation because the gradient function depends on BOTH the x value and y value.

If you only have the x-value, then you have 4 possible answers in this case. However, to find the gradient of the point that Sal plotted, you first need to find its y-coordinate as well using the original function (or u will be given the y-coordinate if you are lucky).

The way you do this is solve the function for the y-value at the given x using a graphing or solving capabilities of calculators or computer programs. You can visually see that the y-coordinate of the point that Sal plottes is POSITIVE and the 2nd GREATEST value, thus, you can input both the x and y coordinates into the gradient function to find the gradient.

I really hope this makes a little sense.(4 votes)

- The applying chain rule to y^2 thing. Is there a reason? Does it always go that you have to use chain rule to get the derivative of something that has y?(2 votes)
- Yes. The whole point of implicit differentiation is to differentiate an
*implicit equation*, that is, an equation that is not explicitly solved for the dependent variable 𝑦. So whenever we come across a 𝑦 term when implicitly differentiating, we must*assume that it is a function of*𝑥. So by assuming it is a function of 𝑥 (without knowing the function explicitly), we differentiate 𝑓(𝑦) as a composition of functions 𝑓 and 𝑦. In this case, 𝑓(𝑥) = 𝑥², so to differentiate 𝑓(𝑦), we note that 𝑦 itself is a function of 𝑥 and we have by chain rule:

[𝑓(𝑦)]' = 𝑓'(𝑦) • 𝑦' = 2𝑦 • 𝑦'

The key is to note that we are always assuming that 𝑦 is a function of 𝑥 when we implicitly differentiate without*explicitly solving for*that function. So whenever we find something "being done to" 𝑦, we basically have a function composition of 𝑦 and the function that describes whatever is being done to it. And of course, to differentiate a composition, you need chain rule!

I hope you understand the logic behind this! Many people just "memorize" the steps to implicit differentiation without understanding why 𝑦 terms are treated the way they are. Feel free to comment if you have questions!(7 votes)

- Would:

dy/dx = (6x*(x^2+y^2) - 10x*y^2) / ( (10*(x^2)**y) - 6*y**(x^2+y^2)^2 )

be equally valid?(2 votes)- Yes, except in the numerator you need to square (x^2 + y^2).

Other than that you could just factor a -1 out of the numerator and denominator to get the same answer that Sal got.(4 votes)

- Are you not able to multiply 6x by (x^2y^2)^2 to get 6(x^3y^2)^2?(2 votes)
- I'm a little confused where you saw (x^2y^2)^2. What I saw was 6x(x^2 + y^2)^2. That plus in there makes a huge difference. Otherwise still no, you have to perform exponents FIRST (PEMDAS). Which requires you take the exponent of (x^2y^2) first, giving you x^4y^4, then you may multiply by 6x, giving you 6x^5y^4.

Regardless I didn't see an (x^2y^2)^2 in the video, I saw (x^2 + y^2) ^2, which is an entirely different story since you must foil that (x^2 +y^2) first, not simply raising the terms to powers and giving the false result of (x^2 + y^2)^2 = x^4 + y^4; <-- that is incorrect since you didn't foil!!(3 votes)

- 1:14Sal derives two things in the second bracket. X^2 with respect to x and Y^2 with respect to why. Can you even do that? same bracket derived with respect to different values?(1 vote)
- Sal derives y^2 with respect to x by the chain rule. Using the chain rule he first derives y^2 with respect to y and then y with respect to x. This is the basic tenet of implicit differentiation. It starts to look a bit hairy and magical when the thing you are deriving gets more complicated. Go back and look at the first lecture on implicit differentiation and that will probably help.(4 votes)

- so if you find the equation, how do you used it to graph out the picture (assuming you dint know what it look like).(2 votes)
- The derivative isn't for the graph of the function, its for the slope at particular points. For the graph you need to take the original function, "(x^2+y^2)^3 = 5x^2y^2", and plug in x-values (or y-values depending on preference) and plot the output as points until you can make a proper graph. Hope this helps!

There are also tricks to find the shape of the graph from the equation alone.(1 vote)

## Video transcript

Once again, I have some crazy
relationship between x and y. And just to get a sense of
what this might look like, if you plot all the x's and y's
that satisfy this relationship, you get this nice
little clover pattern. And I plotted this
off of Wolfram Alpha. But what I'm curious
about in this video, as you might imagine
from the title, is to figure out
the rate at which y is changing with respect to x. And we're going to have
to do it implicitly. We're going to have to find the
implicit derivative of this. We're going to have to derive
this implicitly or take the derivative of it implicitly. So let's apply our
derivative operator to both sides of the derivative
with respect to x on the left and the derivative with
respect to x on the right. So once again, we
apply our chain rule. The derivative of something to
the third power with respect to that something is
going to be three times that something squared. And then we have
to multiply that times the derivative of the
something with respect to x. So the derivative of this
thing with respect to x is going to be 2x,
that's the derivative of x squared with respect to
x, plus the derivative of y squared with respect
to y is going to be 2y times the derivative
of y with respect to x. Once again, we're applying the
chain rule right over here. The derivative of something
squared with respect to the something, which is
2y, times the derivative of the something with
respect to x, which is dy dx. Now, that is going
to be equal to what we have on the right hand side. So we have a 5 times x
squared times y squared. We can take the 5 out
of the picture for now. Take the 5 onto the-- take
it out of the derivative. The derivative of
5 times something is the same thing as 5
times the derivative. And now we can apply
the product rule. So it's going to be 5 times
the derivative of x squared is just going to be
2x times y squared. So that's just the derivative
of the first function times the second function. Plus the first function,
not taking its derivative, x squared times the derivative
of the second function. Well, what's the derivative of
y squared with respect to x? Well, we already figured it out. It's the derivative of
y squared with respect to y, which is 2y times the
derivative of y with respect to x. Let me make it clear
what I just did. This is this, and then this
is when I took its derivative. So that is when I applied
the derivative operator. Similarly, that is that. And when I applied the
derivative operator, I got that. The derivative with respect
to x right over there. So let's see if we can
somehow solve for dy dx. So what I'm going to do
on the left hand side is I'm just going to distribute
this purple thing onto both of these terms. So if you distribute
this purple thing onto this term
right over here, you get 3 times 2x, which is 6x,
times x squared plus y squared, squared. And then if you distribute
this purple stuff on to this one right over
here you get plus, let's see, 2y times 3 is 6y times x
squared plus y squared. Let me make sure. 2y times 3 is 6y times x
squared plus y squared, squared, and then I'll keep
the dy dx in that green color. dy dx is equal to, well,
we can multiply the 5 times this business right over here. And so everything
that's not a dy dx term maybe I will
do in purple now. So you do 5 times
this stuff right over here, which gives
you 10xy squared. And then 5 times all
of this right over here is going to be plus
10x squared y dy dx. Did I do that right? Yep, that looks
just about right. And now we have to
solve for dy dx. What I'm going to
do is I'm going to subtract this 10x squared. I'm going to subtract
the 10x squared y dy dx from both sides. 10x squared y dy dx. Derivative. That's not green. Derivative of y
with respect to x. Going to subtract
that from both sides so that I can get it
on the left hand side. dy dx. And I'm going to subtract
this business, this 6x times all this craziness
from both sides. So minus 6x times x squared
plus y squared, squared. Let me subtract it
from here as well. Minus 6x x squared
plus y squared squared. And what are we left with? Well, these guys cancel out. On the left hand
side right over here, we are left with 6y times
x squared plus y squared, squared minus 10x
squared y times dy dx, the derivative
of y with respect to x. The derivative of
y with respect to x is equal to-- these
characters cancel out-- and we are left with 10xy
squared minus 6x times x squared plus y
squared, squared. And now if we want
to solve for dy dx, we just divide both
sides of this equation by this business
right over here. And you get the derivative
of y with respect to x. And we deserve a drum roll now. The derivative of
y with respect to x is equal to all of this stuff. And I'm just going
to copy and paste it. So let me copy and paste it. All of that stuff
over all of this stuff over all of that stuff. Once again, just going
to copy and paste it. And I just have to draw
the little fraction symbol. So we are done. We have figured out-- and it
was kind of a hairy expression, but it didn't take
us too long-- what the derivative of y with
respect to x at any point is. So what you'd want to do
is if you want to say, hey, what is the slope of
the tangent line right at-- let me do this in a color
that you can actually see. What is the slope of the tangent
line right at that point? Well, you'd want to figure out
what the x-coordinate of that point it so you
could say maybe x is this value right over here. And then you can solve
for y to figure out what the y value would be. And then you would
chunk that x and y into this hairy
expression right over here to figure out the slope
of the tangent line.