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Second derivatives (implicit equations): find expression

Given an implicit equation in x and y, finding the expression for the second derivative of y with respect to x.

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  • male robot hal style avatar for user Yash Vij
    Isn't it possible to rewrite the second derivative as 4 / y^3 because if you create a common denominator you get (y^2 - x^2) / y^3? And since we know from the function that y^2 - x^2 = 4, we can substitute 4.
    (41 votes)
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  • female robot grace style avatar for user why do you care
    I tried to use the quotient rule but I got a different result. Is it me doing it incorrectly or that it has to be done as Sal did?
    What I did was the same except the fact that I used quotient rule, which is as follows:

    1.y - x.1 / y^2
    (6 votes)
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    • cacteye blue style avatar for user Jerry Nilsson
      Remember that we're differentiating with respect to 𝑥, which means that the derivative of 𝑦 is 𝑑𝑦∕𝑑𝑥, not 1.

      So, applying the quotient rule, we get
      𝑑²𝑦∕𝑑𝑥² = (1・𝑦 − 𝑥・𝑑𝑦∕𝑑𝑥)∕𝑦² = 1∕𝑦 − (𝑥∕𝑦²)・𝑑𝑦∕𝑑𝑥

      and since 𝑑𝑦∕𝑑𝑥 = 𝑥∕𝑦, we get
      𝑑²𝑦∕𝑑𝑥² = 1∕𝑦 − (𝑥∕𝑦²)・(𝑥∕𝑦) = 1∕𝑦 − 𝑥²∕𝑦³
      (19 votes)
  • stelly orange style avatar for user Lucas
    At and why do we multiply by the derivative?

    I see that this probably has something to do with it being a "y". But I'm just not really sure why we use it.
    (7 votes)
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    • leaf green style avatar for user kubleeka
      It's because of the chain rule, and because y is a function of x. If we have a function like f(g(x)), the derivative is f'(g(x))·g'(x). This is the same thing, with the outer function f(x)=x² (at anyway) and the inner function g(x)=y(x).
      (10 votes)
  • aqualine seed style avatar for user Pascale
    What if x and y were raised to a fractional power instead of an integer? For example, what if you had x^1/3 + y^1/3 = 4? I know how to take the derivative of x^1/3 (power rule), but I'm having trouble with how I should take the derivative of y^1/3.
    (4 votes)
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  • piceratops tree style avatar for user Michael Wade
    Could you have instead applied the derivative operator to both sides of the equation before solving for dy/dx?

    d/dx[ 2y(dy/dx) - 2x ] = d/dx[0]
    (6 votes)
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    • blobby green style avatar for user Chuck B
      Try it! Yes, you can -- but you end up with a (dy/dx)^2 term in the result. So for this problem, it is advantageous to work out the first derivative dy/dx first.

      In fact, you can even start off the whole problem by taking the second derivative d^2/dx^2 of both sides. You end up in the same place.
      (2 votes)
  • starky ultimate style avatar for user Cabeça Ameixa
    If I solve the equation for y before taking the first derivative, I get the result
    ±( (x) / (√(x² + 4)) )
    Does this result have any meaning?
    (4 votes)
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  • hopper cool style avatar for user Iron Programming
    At sal ways "that we are going to use the chain rule here" and so he takes the derivative of (y^2) with respect to "y" and gets 2y, and then multiplies it by the the derivative of (y^2) with respect to "x" which is just (dy/dx), and then he says that the derivative of (x^2) with respect to "x" is 2x.
    So what I don't get, is how he used the chain rule there, and why doesn't he do the same thing for (x^2)?
    It almost seems like he's changing the value.
    I've seen lot's of examples like this, and I thought I understood, but now I'm not so sure.....
    Please answer, thanks in advance,
    - ncochran2
    (3 votes)
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    • piceratops ultimate style avatar for user ggadget6
      When we do implicit differentiation, we say that one of the variables is a function of the other. In this case, we are saying that y is a function of x. We are looking for dy/dx, which is the derivative with respect to x. To do this, we take the derivative with respect to x of both sides (that's what the d/dx means). We already know the derivative with respect to x of x^2 - It's 2x! There's not need to use the chain rule here because we are taking the derivative with respect to x. However, when we take the derivative of y^2 with respect to, we must use the chain rule. Why? Because y is a function of x! So, we take the derivative of y^2with respect to y first, and then we can multiply it by the derivative of y with respect to x.


      Another thing to note: if we did want to use the chain rule for x^2, you technically could. You take the derivative of x^2 with respect to x, which is 2x, and multiply it by the derivative of x with respect to x. However, notice that the derivative of x with respect to x is just 1! (dx/dx = 1). So, this shouldn't change your answer even if you choose to think about the chain rule.
      (6 votes)
  • duskpin ultimate style avatar for user Cindy Wang
    So when do we know to use the implicit equation and not the traditional methods?
    (3 votes)
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    • female robot grace style avatar for user loumast17
      You kind of use the same method no matter what. When you have an equation you take the derivative of both sides then use algebra to find what dy/dx is.

      USUALLY y is by itself on one side, and the derivative of y is dy/dx, so no algebra is necessary in that case.

      Then once you have dy/dx it's pretty simple to find the second and above derivative. Does that help?
      (4 votes)
  • aqualine tree style avatar for user lucy lee
    i kinda don't understand why i have to treat Y as a VARIABLE whilst taking the derivative of the whole equation with respect to X.
    (3 votes)
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    • piceratops ultimate style avatar for user Learner
      In implicit function, both x and y are used as variables. However, they are not used in the same way x and y are used in explicit functions, where y is entirely dependent upon x. Implicit functions simply map all the points (x,y) in which the function is true. So the function is dependent upon x and y, thus we must treat both like variables.
      (4 votes)
  • orange juice squid orange style avatar for user Evan
    How would I graph these derivatives and second derivatives? Since the equations have both x and y in them, would I have to isolate y before graphing? Or would I simply replace dy/dx with y? Thanks for the help!
    (2 votes)
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    • leaf green style avatar for user kubleeka
      You would isolate dy/dx on the left, and have some expression in x and y on the right. Then you would replace each y with the original expression in x, leaving you with
      dy/dx=(expression in x).

      Then plot dy/dx on the vertical axis and x on the horizontal axis.
      (4 votes)

Video transcript

- [Instructor] Let's say that we're given the equation that y squared minus x squared is equal to four. And our goal is to find the second derivative of y with respect to x, and we want to find an expression for it in terms of x's and y's. So pause this video, and see if you can work through this. All right, now let's do it together. Now, some of you might have wanted to solve for y and then use some traditional techniques. But here, we have a y squared, and so it might involve a plus or a minus square root. And so some of y'all might have realized, hey, we can do a little bit of implicit differentiation, which is really just an application of the chain rule. So let's do that. Let's first find the first derivative of y with respect to x. And to do that, I'll just take the derivative with respect to x of both sides of this equation. And then what do we get? Well, the derivative with respect to x of y squared, we're gonna use the chain rule here. First, we can take the derivative of y squared with respect to y, which is going to be equal to two y, and then that times the derivative of y with respect to x. Once again, this comes straight out of the chain rule. And then, from that, we will subtract, what's the derivative of x squared with respect to x? Well, that's just going to be two x. And then last, but not least, what is the derivative of a constant with respect to x? Well, it doesn't change, so it's just going to be equal to zero. All right, now we can solve for our first derivative of y with respect to x. Let's do that. We can add two x to both sides, and we would get two y times the derivative of y with respect to x is equal to two x. And now I can divide both sides by two y, and I am going to get that the derivative of y with respect to x is equal to x, x over y. Now, the next step is let's take the derivative of both sides of this with respect to x, and then we can hopefully find our second derivative of y with respect to x. And to help us there, actually let me rewrite this. And I always forget the quotient rule, although it might be a useful thing for you to remember. But I could rewrite this as a product, which will help me at least. So I'm going to rewrite this as the derivative of y with respect to x is equal to x times y to the negative one power, y to the negative one power. And now, if we want to find the second derivative, we apply the derivative operator on both sides of this equation, derivative with respect to x. And our left-hand side is exactly what we eventually wanted to get, so the second derivative of y with respect to x. And what do we get here on the right-hand side? Well, we can apply the product rule. So first, we can say the derivative of x with respect to x, well, that is just going to be one times the other thing, so times y to the negative one power, y to the negative one power. And then we have plus x times the derivative of y to the negative one. So plus x, what's the, times, what's the derivative of y to the negative one power? Well, first, we can find the derivative of y to the negative one power with respect to y. We'll just leverage the power rule there. So that's going to be negative one times y to the negative two power. And then we would multiply that times the derivative of y with respect to x, just an application of the chain rule, times dy/dx. And remember, we know what the derivative of y with respect to x is. We already solved for that. It is x over y. So this over here is going to be x over y. And so now we just have to simplify this expression. This is going to be equal to, and I'll try to do it part by part, that part right over there is just going to be a one over y. And then all of this business, let's see if I can simplify that. This negative is going to go out front, so minus, and then I'm going to have x times x in the numerator. And then it's going to be divided by y squared and then divided by another y. So it's going to be minus x squared over y to the third, over y to the third, or another way to think about it, x squared times y to the negative three. And we are done. We have just figured out the second derivative of y with respect to x in terms of x's and y's.