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# Worked example: Implicit differentiation

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

## Video transcript

let's get some more practice doing implicit differentiation so let's find the derivative of Y with respect to X we're going to assume that Y is a function of X so let's apply our derivative operator to both sides of this equation so let's apply our derivative operator and so first on the left hand side we essentially are just going to apply the chain rule first we have some the derivative of the derivative with respect to X of x minus y squared so the chain rule tells us this is going to be the derivative of the something squared with respect to the something which is just going to be two times X minus y to the first power I won't leave I won't write the one right over there times the derivative of the something with respect to X well the derivative of X with respect to X is just 1 and the derivative of Y with respect to X that's what we're trying to solve so it's going to be 1 minus dy 1 minus dy DX let me make it a little bit clearer what I just did right over here this right over here is the derivative of X minus y X minus y squared with respect to X minus y with respect to X minus y and then this right over here is the derivative of X minus y with respect with respect to X just the chain rule now let's go to the right-hand side of this equation this is going to be equal to the derivative of X with respect to X is 1 the derivative of Y with respect to X we're going to write that as the derivative of Y with respect to X and then finally the derivative with respect to X of a constant that's just going to be equal to 0 now let's if we can solve for if we can solve for the derivative of Y the derivative of Y with respect to X so the most obvious thing to do let's make it clear this right over here I can rewrite as 2x minus 2y so let me do that so I can save some space this is 2x minus 2y if I just distribute the 2 and now I can distribute the 2x minus 2i onto each of these terms so 2x minus 2y times 1 is just going to be to X minus 2y and then to X minus 2y times negative dy/dx that's just going to be negative 2x minus 2i or we could write that as 2 y minus 2 x times dy DX times dy/dx is equal to 1 plus dy DX is equal to 1 plus I'll do all my dy/dx is in orange now 1 plus dy DX so now there's a couple of things that we could attempt to do we could subtract 2x minus 2y from both sides so let's do that so let's subtract 2x minus 2y from both sides so over here we're going to subtract minus we're going to subtract 2x minus 2y from that side and then we could also subtract a dy/dx from both sides so that all of our dy DX's are on the left-hand side and all of our non dy DX's are on the right-hand side so let's do that so we're going to subtract a DUI DX on the right and a and a dy/dx here on the left and so what are we left with what are we left with well on the left-hand side these cancel out and we're left with 2y minus 2x dy/dx minus 1 dy DX or just minus the dy DX let me make it clear we could write this as a minus 1 dy DX so this is we can essentially just add these two coefficients so this simplifies to 2y minus 2x minus 1 times the derivative of Y with respect to X the derivative of Y with respect to X which is going to be equal to on this side this cancels out we are left with 1 minus 2x plus 2y so let me write it that way or we could write this as we could write this as so negative negative 2y is just a positive 2y and then we have minus 2x and then we add that one plus 1 and now to solve for dy do X we just have to divide both sides by 2 i minus 2x minus 1 and we are left with we deserve a little bit of a drumroll at this point as you can see the hardest part was really the algebra to solve for dy/dx we get the derivative of y with respect to x is equal to 2y minus 2x plus 1 over 2y minus 2x minus 1
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