Applying implicit differentiation, 2023 Calculus AB free response 6a
About this video
Sal uses implicit differentiation on the curve 6xy = 2 + y^3 to show that dy/dx = 2y/(y^2 - 2x). He applies the product rule to 6xy and the chain rule to y^3, then collects the dy/dx terms, factors and divides.
You'll be able to
- differentiate both sides of an equation implicitly with respect to x.
- use the product rule on a term like 6xy.
- collect dy/dx terms, factor and solve for dy/dx.
Key words
- implicit differentiation
- Taking the derivative of both sides with respect to x when y can't easily be written as a function of x, using the chain rule on y terms.
- product rule
- The derivative of a product: derivative of the first times the second, plus the first times derivative of the second.
- dy/dx
- The derivative of y with respect to x.
Transcript
- [Instructor] We're told to consider the curve given by the equation 6XY is equal to two plus Y to the third. Show that the derivative of Y with respect to X is equal to 2Y over Y squared minus 2X. So like always, pause this video, have a go at this before we do this together.
All right, so when you are dealing with a relation that you can't algebraically manipulate easily to get it in the form of Y is equal to F of X, if we wanna find the derivative of Y with respect to X, we will really need to turn to implicit differentiation, and implicit differentiation, what we're going do is take the derivative with respect to X of each side and use, essentially, the chain rule.
So the derivative with respect to X of, I'm just getting the derivative of both sides of this, of 6X times Y is equal to, and I actually like to do brackets round this, so 6XY is equal to the derivative with respect to X of two plus Y to the third. Now, for the left hand side right over here, I could treat this as the product of two functions, 6X and then Y, so I'm just gonna apply the product rule here.
So this is first gonna be the derivative of 6 X with respect to X, which is six times Y, times Y, actually, let me put this in a higher contrast color, six times Y plus 6X, plus 6X, plus 6X, times the derivative of Y with respect to X. Well, we can just write that as DY, DX, the derivative of Y with respect to X.
That's our left hand side. And then our right hand side, this is all going to be equal to derivative with respect to X of a constant, it's just zero, derivative with respect to x of Y to the third is going to be 3Y squared times the derivative of Y with respect to X, derivative of Y with respect to X.
And now what we wanna do, well, actually, we can divide both sides by three. It'll simplify things a little bit. So this will become a two, this will become a two, that just becomes a one, and now, we can subtract this term from both sides. I want to get all the derivative Ys with respect to X on one side, then factor it out, and then see if I can get to what we need to show.
So the left hand side is 2Y, I'm gonna subtract this from both sides, so this is going to go away. And on the right hand side, I am going to be left with Y squared times the derivative of Y with respect to X minus 2X times the derivative of Y with respect to X.
And now, I can say 2Y is equal to, I'll factor out a derivative of Y with respect to X, DY, DX times Y squared minus 2X, and now, I just divide both sides by Y squared minus 2X, and I'm going to get 2Y over Y squared minus 2X is equal to the derivative of Y with respect to X.
And that's exactly what we wanted to show.
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