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Worked example: Evaluating derivative with implicit differentiation

Sal finds the slope of the tangent line to the curve x²+(y-x)³=28 at x=1 using implicit differentiation. Δημιουργήθηκε από τον Σαλ Καν.

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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.