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Implicit differentiation (advanced example)

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

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  • marcimus pink style avatar for user novamwilliams2
    Why, early in the equation solving, was the 5 taken out of the derivative, , and then multiplied back in later, , 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)
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    • leaf green style avatar for user Lucas Van Meter
      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)
  • blobby green style avatar for user O
    How is the derivative of y^2 with respect to x : 2y (dy/dx)?
    (16 votes)
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    • aqualine ultimate style avatar for user shannon.kollasch
      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)
  • mr pants teal style avatar for user Eli Spina
    Why does Sal keep distributing terms like this at ? 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)
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    • leaf green style avatar for user ArDeeJ
      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)
  • leafers ultimate style avatar for user hiddename
    He could simplify the end expression abit right? atleast divide it by 2.
    Is there a spesific reason he didnt do that?
    (5 votes)
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    • piceratops ultimate style avatar for user Just Keith
      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.
      (11 votes)
  • leaf blue style avatar for user Jeff
    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)
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    • male robot hal style avatar for user Farruh Mavlonov
      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)
  • leafers sapling style avatar for user Marianne
    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)
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    • mr pink red style avatar for user andrewp18
      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)
  • male robot hal style avatar for user R Agmoyri
    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)
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  • leaf green style avatar for user Oskars Sjomkāns
    Sal 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)
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    • leafers ultimate style avatar for user KrisSKing
      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)
  • blobby green style avatar for user Leah Wolfe
    Are you not able to multiply 6x by (x^2y^2)^2 to get 6(x^3y^2)^2?
    (2 votes)
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    • aqualine ultimate style avatar for user stolenunder
      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!!
      (2 votes)
  • blobby green style avatar for user Tien Ngo
    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)
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    • male robot hal style avatar for user Kartik Nagpal
      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.