show that the relation (x^3 + y^3 = 6xy) implicitly defines a solution to (dy/dx)=(2y-x^2)/((y^2)-2x)
Use implicit differentiation, plus product rule and chain rule as you know them. The only difference is that y is a function of x, f(x), so when you take the derivative of y, you must multiply the result by dy/dx. This is just an application of chain rule, but it's a little complicated to explain and time is short. Taking your equation, the derivative of x^3 is just 3x^2, as usual. The derivative of y^3, however, is 3y^2 times dy/dx, by chain rule, as I said above. On the right side, hide the constant (6), then use product rule to take the derivative. Again, any time you take a derivative of y, multiply by dy/dx. Take derivatives of x normally. Okay, now you have a new equation with a couple of dy/dx's in it. Just like solving simple equations, gather all the terms containing dy/dx onto one side, and all the non-dy/dx terms on the other. Factor out dy/dx, then divide to isolate dy/dx. There. You've got dy/dx.
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