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Current time:0:00Total duration:4:50

Second derivatives (implicit equations): evaluate derivative

AP.CALC:
FUN‑3 (EU)
,
FUN‑3.F (LO)
,
FUN‑3.F.1 (EK)
,
FUN‑3.F.2 (EK)

Video transcript

so we have a question here from the 2015 AP Calculus a B test and it says consider the curve given by the equation y to the third minus XY is equal to two it can be shown that the first derivative of Y with respect to X is equal to that so they solve that for us and then Part C of it I skipped Parts A and B for the sake of this video evaluate the second derivative of Y with respect to X at the point on the curve where x equals negative 1 and y is equal to 1 so pause this video and see if you can do that alright now let's do it together and so let me just first write down the first derivative so DUI derivative of Y with respect to X is equal to Y over 3 y squared minus X well if we're concerning ourselves with the second derivative well then we want to take the derivative with respect to X of both sides of this so let's just do that do the derivative operator on both sides right over here now on the left hand side we of course are going to get the second derivative of Y with respect to X but what do we get on the right hand side and there's multiple ways to approach this but for something like this the quotient rule probably is the best way to tackle it I sometimes complain about the quotient rule saying it's just a variation of the product rule but it's actually quite useful in something like this we just have to remind ourselves that this is going to be equal to the derivative of the numerator with respect to X and so that's just going to be derivative of Y with respect to x times the denominator 3y squared minus x minus the numerator y times the derivative of the denominator with respect to X well what's the derivative of this denominator with respect to X well the derivative of 3y squared with respect to X that's going to be the derivative of 3y squared with respect to Y which is just going to be 6y I'm just using the power rule there times the derivative of Y with respect to X all I did just now is I took the derivative of that with respect to X which is the derivative of that with respect to Y times the derivative of Y with respect to X come straight out the chain rule minus the derivative of this with respect to X which is just going to be equal to 1 all of that over remember we're in the middle of the quotient rule right over here all of that over the denominator squared all of that over 3y squared minus x squared now lucky for us they want us to evaluate this at a point as opposed to have to do a bunch of algebraic simplification here so we can say when let me do it over here so when I'll do right here when X is equal to negative 1 and Y is equal to 1 well first of all what's the what's dy/dx going to be the derivative of Y with respect to X let me scroll down a little bit so we have a little bit more space the derivative of Y with respect to X is going to be equal to 1 over 3 times 1 squared which is just 3 minus negative 1 so that's just going to be plus 1 it's going to be equal to 1/4 and so this whole expression over here so I can write the second derivative of Y with respect to X is going to be equal to well we know that that's going to be equal to 1/4 1/4 times 3 times 1 squared which is just 3 minus negative 1 so plus 1 minus 1 so I'll just leave that minus out there times 6 times 1 times 1/4 let me just write it out 6 6 times 1 times 1/4 minus 1 all of that over let's see this is going to be 3 times y squared Y is 1 so this is going to be 3 3 minus negative 1 so plus 1 squared now what is this going to be and this is just simplifying something here 1/4 times 4 that's going to simplify to 1 and let's see this is going to be 1 and 1/2 minus 1 so that's going to be 1/2 and then we're going to have all of that over 16 and so this is going to be equal to get a mini drum roll here this is going to be equal to 1 minus 1/2 which is equal to 1/2 over 16 which is the same thing as 1 over 32 and we are done