respect to x. Recall that every time you take the derivative of y to multiply by -, using the Generalized Power Rule. d ay + xy + 63 dx [xy + 63] = X- dx X y Step 3 Now, we put these together and solve for y by getting all the dyterms on the left side of the equation and factoring. d [ x 2 + y 2 ] = - [xy+ 63] dx 2x + 2 y y = y + X- dy dx dx 2y- dy X- dy - 2x dx dx y - 2x 2y - X dy 2y - I =y - 2X dx dy y - 2x y - dx 2y -X 2y - T Step 4 Finally, we need to evaluate the derivative at (9, 3). Substituting 9 for x and 3 for y in the derivative, we have the following. dy X dx (9 , 3 )dx For the equation, find evaluated at the given values. x2 + y2 = xy+ 63 at x = 9, y=3 Step 1 We want to find _ by implicit differentiation and evaluate at the given point. To do so, we take the derivative dx of both sides of the equation, staring with the left hand side. View y as a function of x and take the derivative with respect to x. Recall that every time you take the derivative of y to multiply by , using the Generalized dy Power Rule. dy dx d[ x2 + y2] = xy +63 y + x X X 2y dx Step 2 Now, we take the derivative of the right hand side. To do so, we note that the function on the right is a product, and hence we must use the Product Rule. View y as a function of x and take the derivative with respect to x. Recall that every time you take the derivative of y to multiply by , using the Generalized Power Rule. d dy [xy + 63] = X- dy + dx xy + 63 X y Step 3 Now, we put these together and solve for - by getting all the terms on the left side of the equation and factoring. d [ x 2 + y 2 ] = - [xy + 63] 2x + 2y- ay = y+xy dx dx