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Worked example: Evaluating derivative with implicit differentiation

AP.CALC:
FUN‑3 (EU)
,
FUN‑3.D (LO)
,
FUN‑3.D.1 (EK)
Sal finds the slope of the tangent line to the curve x²+(y-x)³=28 at x=1 using implicit differentiation. Created by Sal Khan.

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  • blobby green style avatar for user Stephen
    at you decide to distribute, Why? could you just put all non dy/dx terms on the opposite side of the equation and then add 1 to both sides making the final equation dy/dx= (1-2x)/3(y-x)^2
    (2 votes)
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  • piceratops ultimate style avatar for user Justin Scheidler
    How do you know when you have to use implicit differentiation? Is it when you are trying to find the rate of change with respect to another function? So if you wanted to know the rate at which f(y) is changing with respect to x with something like x^2y^2=1-- this is when we use this rule?
    (7 votes)
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  • mr pants teal style avatar for user box 0f rox
    I'm assuming that simplifying fractions for problems like these is a bad idea. My initial instinct was to cancel out the 27s in the numerator and denominator.
    (13 votes)
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  • blobby green style avatar for user Rebecca Kurtz
    At about , why would you not simplify by crossing out the 3(y-x)^2's and having -2x?
    (2 votes)
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  • purple pi purple style avatar for user Darko
    At the end of the video, I could see why he didn't further simplify expression of dy/dx so that he could show us we do not have to substitute y for an explicit expression of the original equation, he wanted to show us, if we know y and we know x, we can just substitute those into the expression of dy/dx. But, he could 've simplified it to 1-(2x/3(y-x)^2 and still made his point. Wouldn't that be much simpler?
    (2 votes)
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  • blobby green style avatar for user Carolyn Cassidy
    Okay, so instead of factoring (3(y-x)^2)((dy/dx)-1) instead I moved subtracted 2x from both sides. then I divided 3(y-x)^2 from both sides.
    That gave me dy/dx -1 = -2x/(3(y-x)^2). Then I added 1 to both sides.
    My final equation looked like: dy/dx = -2x/(3(x-y)^2) -1.

    And as hard as I try, I don't understand why that is wrong. As a matter of fact, it seems much easier than factoring 3(x-y)^2. But, when it came to plugging in the coordinate (1,4) I got myself into a muddle because,

    dy/dx = -2(1) / (3(4-1)^2) - 1
    dy/dx = -2/(3(3)^2) -1
    dy/dx = -2/(3*9) -1
    dy/dx = -2/27 -1
    Obviously that isn't the answer Sal got. So, I was wondering what I am missing and why factoring earlier on changes things so dramatically.
    (2 votes)
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  • blobby green style avatar for user Sec Ar
    I suppose the question is redundant since he did it, but can you just multiply the "3(y-x)^2" in its entirity with a term inside the brackets? I've never seen something like that.
    (3 votes)
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  • blobby green style avatar for user JDM
    When you do d/dx(xy) how does it come out to be y+x*(dy/dx)? I'm confused on what steps you take and how you got the outcome
    (2 votes)
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    • aqualine ultimate style avatar for user famousguy786
      The product rule for differentiation has been used.
      First we differentiate x with respect to x(which is 1)multiplied by y remaining as it is which turns out to be y*1=y.
      This is added to the differentiation of y with respect to x which is clearly the derivative of y dy/dx, which is multiplied by x remaining as it is. That's how we get y+x*(dy/dx). I hope this helps.
      (3 votes)
  • male robot hal style avatar for user GabrielZub
    How would you differentiate xy? I'm getting confused trying to work it out. What is the inner function and the outer function?
    (2 votes)
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    • duskpin ultimate style avatar for user Claire Miller
      First, we note that two functions are multiplied together, meaning that we need to use the product rule. The two functions are the function of x (just x in this case) and the function of y (which is also just y), correct?

      So, the product rule says that the derivative of f*g, where f and g are both functions, is f'*g + f*g'. If we apply this to xy and differentiate with respect to x, we have [(d/dx)x * y] + [x * (d/dx)y].

      The derivative of x is just 1. The derivative of y with respect to x is slightly more complex. Since y is a function of x, the derivative of y with respect to x is dy/dx, or y' (whichever notation you prefer). If we substitute this in, the final result is: y + xy'.

      Hopefully this made sense. If not, feel free to ask any clarifying questions you have. Best of luck as you continue your studies!
      (2 votes)
  • piceratops tree style avatar for user raphael.neuburger
    At why didnt he just plug in the x and y which we already knew and then the algebra would become a lot easier?
    (2 votes)
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    • piceratops ultimate style avatar for user Ronald Waite
      I agree Raphael, and when I solved the problem myself before finishing the video, this is exactly what I did. I feel like it was easier than Sal's solution too, since my next step was just:

      2 + 27(dy/dx) - 27 = 0

      then:
      27(dy/dx) = 25

      and finally:
      dy/dx = 25/27
      (2 votes)

Video transcript

We've been doing a lot of examples where we just take implicit derivatives, but we haven't been calculating the actual slope of the tangent line at a given point. And that's what I want to do in this video. So what I want to do is figure out the slope at x is equal to 1. So when x is equal to 1. And as you can imagine, once we implicitly take the derivative of this, we're going to have that as a function of x and y. So it'll be useful to know what y value we get to when our x is equal to 1. So let's figure that out right now. So when x is equal to 1, our relationship right over here becomes 1 squared, which is just 1 plus y minus 1 to the third power is equal to 28. Subtract 1 from both sides. You get y minus 1 to the third power is equal to 27. It looks like the numbers work out quite neatly for us. Take the cube root of both sides. You get y minus 1 is equal to 3. Add 1 to both sides. You get y is equal to 4. So we really want to figure out the slope at the point 1 comma 1 comma 4, which is right over here. When x is 1, y is 4. So we want to figure out the slope of the tangent line right over there. So let's start doing some implicit differentiation. So we're going to take the derivative of both sides of this relationship, or this equation, depending on how you want to view it. And so let's skip down here past the orange. So the derivative with respect to x of x squared is going to be 2x. And then the derivative with respect to x of something to the third power is going to be 3 times that something squared times the derivative of that something with respect to x. And so what's the derivative of this with respect to x? Well the derivative of y with respect to x is just dy dx. And then the derivative of x with respect to x is just 1. So we have minus 1. And on the right-hand side we just get 0. Derivative of a constant is just equal to 0. And now we need to solve for dy dx. So we get 2x. And so if we distribute this business times the dy dx and times the negative 1, when we multiply it times dy dx, we get-- and actually I'm going to write it over here-- so we get plus 3 times y minus x squared times dy dx. And then when we multiply it times the negative 1, we get negative 3 times y minus y minus x squared. And then of course, all of that is going to be equal to 0. Now all we have to do is take this and put it on the right-hand side. So we'll subtract it from both sides of this equation. So on the left-hand side-- and actually all the stuff that's not a dy dx I'm going to write in green-- so on the left-hand side we're just left with 3 times y minus x squared times dy dx, the derivative of y with respect to x is equal to-- I'm just going to subtract this from both sides-- is equal to negative 2x plus this. So I could write it as 3 times y minus x squared minus 2x. So we're adding this to both sides and we're subtracting this from both sides. Minus 2x. And then to solve for dy dx, we've done this multiple times already. To solve for the derivative of y with respect to x. The derivative of y with respect to x is going to be equal to 3 times y minus x squared minus 2x. All of that over this stuff, 3 times y minus x squared. And we can leave it just like that for now. So what is the derivative of y with respect to x? What is the slope of the tangent line when x is 1 and y is equal to 4? Well we just have to substitute x is equal to 1 and y equals 4 into this expression. So it's going to be equal to 3 times 4 minus 1 squared minus 2 times 1. All of that over 3 times 4 minus 1 squared, which is equal to 4 minus 1 is 3. You square it. You get 9. 9 times 3 is 27. You get 27 minus 2 in the numerator, which is going to be equal to 25. And in the denominator, you get 3 times 9, which is 27. So the slope is 25/27. So it's almost 1, but not quite. And that's actually what it looks like on this graph. And actually just to make sure you know where I got this graph. This was from Wolfram Alpha. I should have told you that from the beginning. Anyway, hopefully you enjoyed that.