 Quandaries and Queries Name: Emily I am: A student Grade Level: 12th grade   Hello :) I am confused with what I am suppose to do for the following problem...   If x^3+3xy+2y^3=17, then in terms of x and y, dy/dx = a. -(x^2+y/x+2y^2) b. -(x^2+y/x+y^2) c. -(x^2+y/x+2y) d. -(x^2+y/2y^2) e. -(x^2)/(1+2y^2)   I'm not sure what to do, but here is what I have so far. Let me know if I am on the right track and what I should do.   x^3+[3(1)+y(3)]+(6y^2)=0 3x^2+3x+3y+6y^2 = x^2+x+y+2y^2   Thank you!   ~Emily Hi Emily, In this problem you need to keep in mind that y is a function of x, that is y is just the name for some expression in x that you don't know. It might be that y = x3 + 4 or y = x2 + 1, you don't know. Hence, when you have to find the derivative of y you can't calculate it, all you can say is that it is the derivative of y with respect to x, that is dy/dx. In particular part of your expression is 3xy and you have to differentiate this. It is a product so you need to use the product rule. So differentiating 3xy I get 3 (the derivative of x) y + 3 x (the derivative of y) the derivative of x is 1 so this is 3 (1) y + 3 x (the derivative of y) The derivative of y you can't calculate so the best you can get is 3 y + 3 x dy/dx Likewise with the term 2y3. The derivative is 2 3 y2 (the derivative of y) = 6 y2 dy/dx Thus with the expression x3 + 3xy + 2y3 = 17 when you differentiate both sides with respect to x you get 3x2 + 3y + 3x dy/dx + 6y2 dy/dx = 0 Now solve foy dy/dx. Penny Go to Math Central