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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)

## Video transcript

we've been doing a lot of examples where we just take implicit different and where we just take implicit derivatives but we actually haven't calculate 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 I want to figure out the slope the slope at X is equal to 1 so when X is equal to 1 and 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 -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 let's skip down here past the orange so the derivative with respect to X of x squared is going to be 2 X is going to be 2 X and then the derivative with respect to X of something to the third power is going to be 3 x that's something it's 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 now we need to solve for dy/dx so we get 2x and so if we distribute if we distribute this business times the dy/dx and times the negative 1 when we multiply it times the D dy/dx we get in lecture I'm going to write it over here so we get plus plus 3 times y minus x squared times dy DX dy DX and then we multiply it times the negative 1 we get negative 3 times y minus X Y minus x1 be careful why 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 gonna write in green so the left-hand side were just left with 3 times y minus x squared times dy DX dy 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 plus sorry 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 could leave it just like that for now so what is the derivative of Y with respect X what is the slope of the tangent line what X is 1 and Y is equal to 4 well we just have to substitute substitute X is equal to 1 and white for into this expression so it's going to be equal to 3 times 4 minus 1 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 squared 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 then 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 well from Wolfram Alpha I should have told you that from the beginning anyway hopefully you enjoyed that
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