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Differentiating related functions intro

Sometimes we have an equation that relates to functions of the same variable. Using the chain rule, we can find the derivatives of those functions with respect to that variable.

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  • starky sapling style avatar for user 20leunge
    I'm so confused, where did t come from? How do we know x and y are functions of t?
    (12 votes)
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  • piceratops ultimate style avatar for user Matthew Chen
    Could you also solve this equation using implicit differentiation?
    (8 votes)
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    • aqualine ultimate style avatar for user Emma
      Using implicit differentiation:
      y=sqrt(x)

      Take the derivative of both sides (note that we are taking dy/dt, not dy/dx, because we are taking the derivative in terms of t as the question calls for):
      dy/dt = (1/2 x^(-1/2))(12)
      where (1/2 x^(-1/2)) is dy/dx and 12 is, as given, dx/dt.

      When dy/dx is multiplied with dx/dt, we get dy/dt.

      Since we are finding dy/dx when x is 9, we get:
      dy/dt = (1/2 (sqrt9)^(-1/2))(12)
      dy/dt = (1/2 * 1/3)(12)
      dy/dt = (1/6)(12)
      dy/dt = 2

      Basically, this is what is shown in the video, but in a more direct format.
      (4 votes)
  • blobby green style avatar for user jeremy.yuan1
    I don't understand how dy/dx becomes (1/2)x^(-1/2) instead of 1/sqrtx
    (3 votes)
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    • female robot grace style avatar for user loumast17
      sqrt(x) can be written as x^(1/2)
      d/dx x^n = n*x^(n-1). Here n =1/2 so let's plug in.

      (1/2)*x^(-1/2) negative exponents can be written as x^(-a) = 1/(x^a) so we can rewrite the answer we got as this

      (1/2)*1/(x^(1/2)) then as we said before x^1/2 can be written as sqrt(x)

      (1/2)*1/sqrt(x) then we just multiply the fractions

      1/(2sqrt(x))

      so 1/sqrt(x) is actually pretty close, just need to keep in mind it is also being multiplied by 1/2

      Let me know if that did not help.
      (4 votes)
  • piceratops ultimate style avatar for user Garret Cervantez
    Can someone tell me if I understand this correctly in regards to dx/dt. After re-watching the video I am less clear than a I was a few minutes ago.

    dx/dt is saying what is the derivative for the FUNCTION x when you insert the PARAMETER t.

    Thanks for the help.
    (2 votes)
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  • winston default style avatar for user J.R. Foster
    Where did the dx/dt come from? How does that relate to the chain rule?

    If we did the chain rule on sqrt(x), that would come out as 1x^-1/2 and that would be it, wouldn't it? Why tack on this extra dx/dt and why is it there?

    Does it have something to do with time being a factor in this equation?
    (3 votes)
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    • female robot grace style avatar for user loumast17
      Sal mentions that the problem states that x AND y are differentiable funtions, so x is also a differentiable function, which means x is a function. the problem then says dx/dt is 12 so that is basically giving us the answer that x's independent variable is t. so you can think of y as y(x) or y of x and x as x(t) or x of t.

      it relates to the chain rule because we are saying y is a function of and x is a function of t so it is one function inside of another, which is what the chain rule deals with.

      We don't use the chain rule here as we normally do. instead we fill in the blanks. the chain rule says dy/dt = dy/dx * dx/dt. we don't know what x(t) is though so we cannot solve for dx/dt, but the problem gives us what it is so we can just fill it in.

      time being a factor is often why you would have a problem like this, but not always, the big takeaway is just how to take derivatives of composite functions that are not written out in the standard way.

      Let me know if something didn't make sense, or yous till have questons.
      (2 votes)
  • starky sapling style avatar for user Shivam
    Shouldn't we have 2answers for this question i.e. 2, -2
    Explanation:

    we want to calculate dy/dt for x= 9 and we know x-y relation so we get y = +3,-3 for which we have to calculate dy/dt

    since y = x^.5 , so x= y^2

    given is, dx/dt = 12
    we substitute x with y^2 so above equation becomes d(y^2)/dt = 12
    so, applying chain rule and simplifying we get,
    dy/dt = 6/y

    substitute two values of y( which we found at top) in this equation. we get
    dy/dt = +2,-2
    (2 votes)
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  • aqualine ultimate style avatar for user Liang
    i used implicit differentiation and solved the problem. i don't get what's new and important being taught that we need to be concerned about learning in this video.
    (1 vote)
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  • mr pink green style avatar for user Yaotian Zhang
    I got different answer somehow.
    dx/dt = 12, get x(t)=12t,
    then when x = 9, t is 3/4.
    y = root x, so y is root of 12t,
    so dy/dt is 1/2*(12t)^(-1/2), plug in t=3/4, the answer is 1/6.
    Please tell me where I did wrong.
    (1 vote)
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    • cacteye blue style avatar for user Jerry Nilsson
      𝑑𝑥∕𝑑𝑡 = 12 ⇒ 𝑥 = 12𝑡 + 𝐶
      (𝐶 will cancel out with itself later on, but for now we should have it there)

      𝑥 = 9 ⇒ 12𝑡 + 𝐶 = 9 ⇒ 𝑡 = (9 − 𝐶)∕12

      𝑦 = √𝑥 = √(12𝑡 + 𝐶)
      ⇒ 𝑑𝑦∕𝑑𝑡 = 1∕(2√(12𝑡 + 𝐶))⋅12 (chain rule!)
      = 6∕√(12𝑡 + 𝐶)

      Plugging 𝑡 = (9 − 𝐶)∕12, we get
      𝑑𝑦∕𝑑𝑡 = 6∕√(12(9 − 𝐶)∕12 + 𝐶) = 6∕√(9 − 𝐶 + 𝐶) = 6∕√9 = 6∕3 = 2
      (1 vote)
  • old spice man blue style avatar for user carter
    I just got confused ! dy/dt = dy/dx.dx/dt right ! and dx/dt = 12 ! plus y= square root of x which means dy/dx = dx/dt over the root of x * 2, which leads to dy/dt = 1/6 am I wrong ?
    (1 vote)
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  • leaf green style avatar for user david powell
    What does "x=f(t)" look like when it's graphed? I tried on Desmos but I just get an error. Is t one of the axes on the plane? If so, which one? Thanks.
    (1 vote)
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    • male robot donald style avatar for user Venkata
      We don't know its graph because x(t) isn't explicitly defined, but you could draw a graph for it on the x-t plane if the equation is given.

      In most cases, you won't be asked to graph it. t here can be considered as a parameter, which is something both x and y depend on. Most people consider it as time, and x(t) and y(t) are seen as the x and y coordinate of a particle w.r.t time. So, on an x-y plane, you can't actually represent t. Instead, you can take x(t) and y(t) and plot points for different t values. The plot you get is the path the particle traces as time increases.
      (1 vote)

Video transcript

- [Instructor] We are told the differentiable functions X and Y are related by the following equation, Y is equal to square root of X. And it's interesting, they're telling us that they are both differentiable functions, even X is a function, must be a function of something else. Well, they tell us that the derivative of X with respect to T is 12 and they want us to find the derivative of Y with respect to T when X is equal to nine. So, let's just make sure we can understand this. So, they're telling us that both X and Y are functions. Arguably they are both functions of T. Y is a function of X but then X is a function of T so Y could also be a function of T. One way to think about it is if X is equal to F of T, then Y is equal to the square root of X which would just be F of T. Another way to think about it if you took T as your input into your function F, you're going to produce X and then if you took that as your input into the square root function you are going to produce Y. So, you could just view this as just one big box here, that Y is a function of Y but now let's actually answer their question. To tackle it we just have to apply the chain rule. The chain rule tells us that the derivative of Y with respect to T is going to be equal to the derivative of Y with respect to X times the derivative of X with respect to T. So, let's apply it to this particular situation. We're gonna have the derivative of Y with respect to T is equal to the derivative of Y with respect to X. Well, what's that? Well, Y is equal to the principle root of X. You could also write this as Y is equal to X to the one half power. We could just use the power rule. The derivative of Y with respect to X is one half X to the negative one half. So, let me write that down. One half X to the negative one half and then times the derivative of X with respect to T times the derivative of X with respect to T. Well, let's see, we wanna find what we have here in orange, that's what the questions asks us. They tell us when X is equal to nine and the derivative of X with respect to T is equal to 12. So, we have all of the information necessary to solve for this. So, this is going to be equal to one half times nine to the negative one half, nine to the negative one half, times DX/DT, the derivative of X with respect to T is equal to 12, times 12. So, let's see, nine to the one half would be three, nine to the negative one half would be 1/3, so this is 1/3, so this will all simplify to 1/2 times 1/3 is 1/6, so we could have a six in the denominator and then we are going to have a 12 in the numerator. So, 12/6, so the derivative of Y with respect to T when X is equal to nine and derivative of X with respect to 12 is two.