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# Showing explicit and implicit differentiation give same result

AP.CALC:
FUN‑3 (EU)
,
FUN‑3.D (LO)
,
FUN‑3.D.1 (EK)

## Video transcript

what I want to show you in this video is that implicit differentiation will give you the same result as I guess we could say explicit differentiation when you can differentiate explicitly so let's say that I have the relationship x times the square root of y is equal to 1 this one is actually pretty straightforward to define explicitly in terms of X to solve for y so if we divide both sides by X we get square root of Y is equal to 1 over X and then if you square both sides you get Y is equal to 1 over x squared which is the same thing as X to the negative 2 power and so if you want the derivative of Y with respect to X this is pretty straightforward this is just an application of the chain rule we get dy/dx is equal to negative 2 X to the negative 2 minus 1 X to the negative 3 power so that's pretty straightforward but I want to see is if we get the same exact result when we differentiate implicitly so let's apply our derivative operator let's apply our derivative operator to both sides to both sides of this equation and so let me make it clear what we're doing x times the square root of y + 1 right over there when you apply the derivative operator to the expression on the left hand side we are just going to have to apply well after we're going to apply both the product rule and the chain rule the product rule tells us so we have the product of two functions of X you could view it that way so this the product rule tells us this is going to be the derivative with respect to X of x of x times the square root of Y times the square root of y plus plus X not taking its derivative plus x times the derivative with respect to x times the derivative with respect to X of the square root of Y let make it clear this bracket of the square root of Y of the square root of Y and on the right hand side right over here the derivative with respect to X of this constant that's just going to be equal to that's just going to be equal to zero so what does this simplify to well the derivative with respect to X of X is just 1 so we're just going to be left this simplifies to 1 so we're just going to be left with the square root of Y right over here so we're just going to be this is going to simplify to a square root of Y and what does this over here simplify to well the derivative with respect to X of the square root of Y here we want to apply the chain rule so let me make it clear so we have plus this X plus this X plus whatever business this is and I'm going to do this I'm going to do this in blue well it's going to be the derivative of the square root of something with respect to that something well the derivative of the square root of something with respect to that something or the derivative of something to the 1/2 with respect to that something is going to be 1/2 times that something to the negative 1/2 power once again this right over here is the derivative of the square root of Y with respect with respect to Y we've seen this multiple times if I were to say the derivative of the square root of x with respect to X you would get you would get 1/2 X to the negative 1/2 now I'm just doing it with wise but we're not done yet remember our derivative operator wasn't the same with respect to Y it's with respect to X so this only gets us with respect to Y we'd apply the entire chain rule we have to multiply that we have to multiply that times the derivative of Y with respect to X in order to get the real derivative of this this expression with respect to X so let's multiply times the derivative of Y with respect to X we don't know what that is that's actually what we're trying to solve for but to use the chain rule we just have to say it's the derivative of the square root of Y with respect to Y times the derivative of Y with respect to X this is the derivative of this thing with respect to X so we get this on the left hand side on the right hand side we just have a 0 and now once again we can attempt to solve for the derivative of Y with respect to X and maybe the easiest first step is to subtract the square root of Y from both sides of this equation and actually let me let me move all of this stuff over so I have once again more room to work with so let me let me let me cut it actually and then let me paste it let me move it over right over here so we went from there to there I didn't I didn't gain a lot of real estate but hopefully this helps a little bit and actually I don't even like that let me put it let me leave it where it was before so then we if we square if we subtract the square root of Y from both sides we get and I'll try to simplify as I go we get this thing which I can rewrite as as X as x times well it's just going to be X in the numerator divided by divided by 2 times the square root of Y Y to the negative 1/2 is just the square root of Y in the denominator and 1/2 I just put the 2 in the denominator there times dy/dx times the derivative of Y with respect to X is going to be equal to is equal to the negative square root of Y I just subtracted the square root of Y from both sides and actually this is something that I might actually want to copy and paste up here so copy and then paste so let's go back up here just to continue our simplification solving for dy/dx well to solve for dy/dx we just have to divide both sides by x over 2 times the square root of y so we're left with we are left with dy DX dy DX is equal to or if we were dividing both sides by this is the same thing as multiplying by the reciprocal of this is equal to 2 times the square root of y over x over my yellow x x times the negative square root of Y well what's this going to simplify to this is going to be equal to Y times the square root of Y times the square root of Y is just Y the negative times the 2 you get negative 2 so you get negative 2y over X is equal to the derivative of Y with respect to X now you might be saying look we just figured out the derivative implicitly and it looks very different than what we did it looks very different than what we did right over here when we just use the power rule we got negative 2 X to the negative third power X and X to the negative 3 power the key here is to realize that this this thing right over here we could solve explicitly in terms of we could solve for y so we could just make this substitution back here to see that these are the exact same thing so if we make the substitution Y is equal to 1 over x squared you get dy/dx the derivative of Y with respect to X is equal to is equal to our negative 2 is equal to the negative 2 times 1 over x squared and then all of that over X which is equal to negative 2 over X to the third which is exactly what we have over here negative 2 X to the negative third power
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