in this video we're going to focus on problems associated with implicit differentiation so given the equation x squared plus y squared is equal to 100 find dydx now i want to go over a few things let's say if you differentiate x cubed with respect to x you know the answer is 3x squared but what about if you differentiate y cube with respect to x it's going to be three y squared times d y d x and for instance if you differentiate let's say r to the fourth power with respect to x it's gonna be four cubed times dr dx for this example dx over dx will cancel so there's no point in writing it so keep that in mind so when doing implicit differentiation we're going to differentiate this function with respect to x so every time you differentiate a y variable you need to add the term dydx to it so let's go ahead and differentiate both sides with respect to x the derivative of x squared is 2x and the derivative of y squared with respect to x is 2y times dydx and the derivative of a constant is zero so now let's isolate d y d x so let's take the two x and move it to the other side so if we subtract both sides by two x on the left we'll have two y 2y dydx and on the right negative 2x now let's divide both sides by 2y we could also cancel negative 2. so d y over d x is equal to negative x over y so now that we have d y d x we can move on to the second part of the problem calculate the slope at the point six comma eight so let's evaluate d y d x at that point so x is six and y is eight so we need to reduce this fraction six is three times two and eight is four times two so we could cancel a two and so d y d x is equal to negative three over four at the point six comma eight and that's the answer now let's move on to number two given the equation x cubed plus 4xy plus y squared is equal to 13 find dydx so let's differentiate everything with respect to x so the derivative of x cubed that's going to be three x squared now what about the derivative of four x y because we need to use the product rule so let's treat for x as if it's f and y as if it's g so here's the form of the product rule the derivative of f times g is going to be the derivative of the first part f prime times the second plus the first part times the derivative of the second part so the derivative of four x is four and we're going to leave the second part the way it is that's g and then plus f the first part times g prime the derivative of the second part the derivative of y is one but with respect to x is going to be one times d y d x and the derivative of y squared is going to be two y d y d x and the derivative of the constant 13 is zero now we need to isolate d y d x so what i'm going to do is out of these two terms i'm going to take out d y d x i'm going to have 4x plus 2y now every term that doesn't have a dydx i'm going to move it to the other side of the equation so i'm going to move 3x squared plus 4y to the right side they're positive on the left side but they're going to be negative on the right side so i have negative 3x squared minus 4y now i need to divide both sides by 4x plus 2y and so now i have the answer to the first part of the problem so dydx is equal to negative 3x squared minus 4y divided by 4x plus 2y now there's nothing else that we can do here to simplify this expression so all we can do is evaluate it at the point one comma two so x is one y is two so one squared is one times negative three four times two is eight and then two times two is four so negative three minus eight is negative eleven four plus four is eight so dydx is equal to negative 11 divided by 8 at the point 1 comma 2. number 3 find dydx by implicit differentiation so let's differentiate both sides with respect to x on five minus x squared is equal to sine and then x y squared so the derivative of five is 0 and the derivative of negative x squared that's negative 2x the derivative of sine is cosine now according to chain rule we need to keep the inside function the same and then we need to multiply by the derivative of the inside so that is the derivative of x y squared so we need to use the product rule so the derivative of the first part x is one times the second y squared plus the first part x times the derivative of the second part the derivative of y squared is two y times d y d x so we need to isolate d y d x well the first thing i'm going to do is i'm going to divide both sides by cosine x y squared actually you know what i'm not going to do that i'm going to distribute it i'm going to distribute cosine to what i have here so negative 2x is equal to y squared cosine xy squared and then plus 2 x y cosine x y squared times d y d x so i'm going to take this term move it to that side so i have negative 2x minus this term and that's equal to 2xy cosine times dydx now the last thing i need to do is divide both sides by 2xy cosine so the final answer is negative 2x minus y squared cosine xy squared divided by 2xy cosine and so that's how you can use implicit differentiation with trigonometric functions now let's try it one more problem let's say if we have x cubed plus y cube is equal to let's say nine find d squared y over dx squared and evaluate it at the point one comma two go ahead and try that so first we need to find the first derivative so let's differentiate both sides with respect to x so the derivative of x cubed is 3x squared the derivative of y cubed is three y squared times d y d x and the derivative of a constant is zero now let's subtract both sides by three x squared so on the right side it's negative three x squared and then let's divide both sides by three y squared so d y d x is going to be negative three divided by three is negative one so d y d x is negative x squared over y squared so we have the first derivative at this point we need to find a second derivative so what we need to do is differentiate this function so the derivative of d y d x with respect to x is going to be d squared y over dx squared and then we need to use the quotient rule for negative x squared over y squared so the derivative of f over g is going to be g f prime minus f of g prime over g squared so in this example f is negative x squared g is y squared so f prime is going to be negative 2x and g prime is 2y dydx so let's use this formula so let's start with g which is y squared times f prime so that's negative two x minus f which is negative x squared times g prime so that's two y d y over d x all divided by g squared which is y squared squared now keep in mind d y d x is negative x squared over y squared so we can replace that with negative x squared over y squared so the second derivative is going to be we can write this as negative two x y squared and then these two negative signs will become positive so we have plus 2 x squared y and then we're going to replace d y d x with negative x squared over y squared and then y squared squared is y to the fourth at this point we could cancel a y so we have negative two x y squared and then two x squared times negative x squared that's going to be minus two x to the fourth power and we still have a y left over on the bottom all divided by y to the fourth now to eliminate the complex fraction let's multiply the top and the bottom by y so we're gonna have negative two x y to the third minus two x to the fourth power over y to the fifth power now if we want to we can take out a negative two x so the second derivative is going to be negative 2x and then we're going to have y cubed and then if we take out negative 2x from this term negative 2x to the fourth divided by negative two x the negative twos will cancel four minus one is three so that's going to give us x cubed all divided by y to the fifth power now we have the point one comma two so now that we have the second derivative in its simplified form we can evaluate it at one two so let's replace x with one and y with two so we have negative two times one and then two to the third is eight one cube is one two to the fifth power that's two times two which is four times two that's eight times two sixteen times two is thirty two eight plus one is nine and thirty two is sixteen times two so we could cancel a two and so the final answer is negative nine over sixteen so that's the value of the second derivative at the point one comma two you