Question

If x^3+y^3−3xy^2=17, find dy/dx in terms of x and y.

Answer 1

Ok so it's been while since I've done this so bear with me.

Answer 2

I believe it is asking you to do the following:
What is the derv of x^3?
What is the derv of y^3? Whatever you get, you also need to multiply a dy/dx to it as well.

Answer 3

Next is the -3xy^2

Answer 4

So you have to treat this almost like the product rule.
So when we are taking the derivative with respect to x we get, -3y^2 (since the dervative of x=1). Then we need to take the derivative with respect to y. So -3x(2y), since derv of y^2 =2y.

Answer 5

Then the derv of 17 will be 0 since 17 is a constant.

Answer 6

Now we need to combine everything we just learned.

Answer 7

So our equation should look like (keeping in mind that derv of x^3=3x^2, derv of y^3=3y^2dx/dy, derv of -3xy^2 with respect to x is -3y^2, derv of -3xy^2 with respect to y is -3x2y*dy/dx [I forgot to add that in]), we have:

Answer 8

3x^2+3y^2(dx/dy)-3y^2-3x2y*(dy/dx) =0

Answer 9

So now you need to get everything that is not (dy/dx) on the right side and leave dy/dx on the left solve (you're solving for dy/dx)

Answer 10

side*

Answer 11

Does that make sense? I know that is a lot of information to throw at you.