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Main content
Current time:0:00Total duration:4:47
AP.CALC:
FUN‑3 (EU)
,
FUN‑3.F (LO)
,
FUN‑3.F.1 (EK)
,
FUN‑3.F.2 (EK)

Video transcript

let's say that we're given the equation that y squared minus x squared is equal to four and our goal is to find the second derivative of Y with respect to X and we want to find an expression for it in terms of X's and Y's so pause this video and see if you can work through this alright now let's do it together now some of you might have wanted to solve for y and then use some traditional techniques but here we have a y squared and so it might involve a plus or minus square root and so some of y'all might have realized hey we can do a little bit of implicit differentiation which is really just an application of the chain rule so let's do that let's first find the first derivative of Y with respect to X and to do that I'll just take the derivative with respect to X of both sides of this equation and then what do we get well the derivative with respect to X of Y squared we're gonna use the chain rule here first we can take the derivative of Y squared with respect to Y which is going to be equal to 2y and then that times the derivative of Y with respect to X once again this comes straight out of the chain rule and then from that we will subtract what's the derivative of x squared with respect to X well that's just going to be 2x and then last but not least what is the derivative of a constant with respect to X well it doesn't change so it's just going to be equal to zero alright now we can solve for our first derivative of Y with respect to X let's do that we can add 2x to both sides and we would get 2y times the derivative of Y with respect to X is equal to 2x and now I can divide both sides by 2y and I am going to get that the derivative of Y with respect to X is equal to X x over Y now the next step is let's take the derivative of both sides of this with respect to X and then we can hopefully find our second derivative of Y with respect to X and to help us there actually let me rewrite this and I'm I always forget the quotient rule although it might be a useful thing for you to remember but I could rewrite this as a product which will help me at least so I'm going to rewrite this as the derivative of Y with respect to X is equal to x times y to the negative one power y to the negative one power and now if we want to find the second derivative we apply the derivative operator on both sides of this equation derivative with respect to X and our left hand side is exactly what we eventually wanted to get so the second derivative of Y with respect to X and what do we get here on the right hand side well we can apply the product rule so first we can say the derivative of X with respect to X well that is just going to be 1 times the other thing so times y to the negative 1 power y to the negative 1 power and then we have plus x times the derivative of Y to the negative 1 so plus X what's that times what's the derivative Y to the negative 1 power well first we can find the derivative of Y to the negative 1 power with respect to Y we'll just leverage the power rule there so that's going to be negative 1 times y to the negative 2 power and then we would multiply that times the derivative of Y with respect to X just an application of the chain rule times dy/dx and remember we know what the derivative of Y with respect to X is we already solved for that it is x over y so this over here is going to be x over Y and so now we just have to simplify this expression this is going to be equal to and I'll try to do it part by part that part right over there it's just going to be a 1 over Y and then all of this business let's see if I can simplify that this negative is going to go out front so minus and then I'm going to have x times X in the numerator and then it's going to be divided by Y squared then divided by another Y so it's going to be minus x squared over Y to the third over Y to the third or another way to think about it x squared times y to the negative three and we are done we have just figured out the second derivative of Y with respect to X in terms of X's and Y's
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