Verified answer

First step would be to try and make the equation look "prettier" by basically moving things around algebraically to remove the square root on y.

1. Multiply all three terms by 3x to get 3(x^2)+y^(1/2)=6x

2. Then move the 3x^2 to the other side to get all x's on one side and the y isolated on the other. This will get you y^(1/2)=6x-3(x^2)

3. Now, you can square the y and square the other side, but do not FOIL the x's out to get y=(6x-3x^2)^2

The reason for that is that you can now differentiate implicitly by using the chain rule and saving some time at the end.

When you differentiate you should get...

dy/dx=2(6x-3x^2)*(6-6x)

If you don't know the chain rule, what I did was first move the power down to the front as a coefficient and subtracted one from the power like you would doing any other derivative. Then I multiplied that by the derivative of whats inside the parentheses.

That is the final answer, but your teacher will probably want you to simplify:

1. By looking at (6x-3x^2), you can factor out a 3x to get 3x(1-x).

2. Also, you can factor out a 6 from the (6-6x) to get 6(1-x)

3. Thus: dy/dx=2[3x(1-x)*6(1-x)]

4. Then you would get 36x(1-x)(1-x)

Finally:

dy/dx=36x(1-x)^2

convey down the means of each x term. And multiply it by utilising the coefficient. Then minus the means term by utilising one. additionally be conscious, x^0 = a million, so 2 = 2*a million = 2*x^0 (replace this in for 2 to tell apart) So, y = 3x^2 - 2x^0 + x^3 y' = 2*3x^(2-a million) - 0*2x^(0-a million) +3*x^(3-a million) y' = 6x^a million - 0 + 3x^2 y' = 6x + 3x^2

You COULD get the √y by itself, square both sides to get rid of the radical, then take the derivative normally. But if you're asked to do this by "implicit differentiation", then taking the derivative of each term with respect to x gives

1 + [ (√y)(-(3x)^-2 *3) + (1/2)y^(-1/2) y' * (3x)^-1 ] = 0

1 + [ -27(√y)(x^-2) + (1/2)y^(-1/2) y' * (3x)^-1 ] = 0

-27(√y)(x^-2) + y^(-1/2) y' * (6x)^-1 ] = -1

y^(-1/2) y' * (6x)^-1 ] = -1 + 27(√y)(x^-2)

y' = [-1 + 27(√y)(x^-2)] * [ 6x√y ]

x +(y^1/2)(3x)^-1 = 2

1 + (1/2)y^(-1/2) * y' * (3x)^(-1) + (y^(1/2)) * -(3x)^(-2) * 3 = 0

y' * (1/2)y^(-1/2) * (3x)^(-1) = (3x)^(-2) * 3 * (y^(1/2)) - 1

y' = [3(3x)^(-2) * (y^(1/2)) - 1] / [(1/2)y^(-1/2) * (3x)^(-1)]