Find dy/dx by implicit differentiation for x^2 - 2xy + y^3 = C

To find dy/dx by implicit differentiation for the given equation x^2 - 2xy + y^3 = C, we'll need to differentiate both sides of the equation with respect to x while treating y as a function of x.

Let's start by differentiating the equation term by term:

d/dx(x^2) - d/dx(2xy) + d/dx(y^3) = d/dx(C)

The derivative of x^2 with respect to x is simply 2x.

For the term 2xy, we need to use the product rule. The product rule states that if we have a function h(x) = f(x)g(x), then h'(x) = f'(x)g(x) + f(x)g'(x).

In this case, f(x) = 2x and g(x) = y. Differentiating f(x) with respect to x gives us f'(x) = 2.

Differentiating g(x) with respect to x gives us g'(x) = dy/dx.

Using the product rule, the derivative of 2xy with respect to x is therefore 2x(dy/dx) + 2y.

For the term y^3, we need to use the chain rule. The chain rule states that if we have a function h(x) = f(g(x)), then h'(x) = f'(g(x)) * g'(x).

In this case, f(y) = y^3 and g(x) = y. Differentiating f(y) with respect to y gives us f'(y) = 3y^2.

Differentiating g(x) with respect to x gives us g'(x) = dy/dx.

Using the chain rule, the derivative of y^3 with respect to x is therefore 3y^2(dy/dx).

Lastly, the derivative of C with respect to x is 0 since C is a constant.

Combining these results, we can rewrite the differentiated equation:

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

Now, let's group the terms involving dy/dx:

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

Factoring out dy/dx:

dy/dx(-2x - 3y^2) = -2x

Finally, we divide both sides by (-2x - 3y^2) to solve for dy/dx:

dy/dx = -2x / (-2x - 3y^2)

Hence, the result is dy/dx = -2x / (-2x - 3y^2).