Calculus question! Calculate dy/dz when xy^3+y=3x ??

Question

I got 3-y^3/1+xy^2. is this correct?

Answer 1

xy^3 + y = 3x

Differentiating implicitly with respect to x, we have to use the product rule on xy^3 and the chain rule on y (to get dy/dx):

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

Move everything without a dy/dx on the right hand side.

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

Factor dy/dx on the left hand side, to get

(dy/dx) [3xy^2 + 1] = 3 - y^3

Divide both sides by (3xy^2 + 1) to get

dy/dx = (3 - y^3) / [3xy^2 + 1]

Seems like I got a different answer than you.

The 3xy^2 comes from the second part of the product rule of differentiating xy^3.

Applying the product rule to xy^3, where x and y^3 are the ones used in the product, we get

x'(y^3) + x(y^3)'

[I'm using the apostrophe to denote taking the derivative of]

x' = 1 (the derivative of x is 1) The derivative of y^3 is 3y^2(dy/dx)

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

And that's why there's a 3xy^2 there to begin with.

Answer 2

Do you mean dy/dx?

If so, then use implicit differentiation (i.e. take the derivative of both sides with respect to x while taking into account that y is still a function of x).

x(3y^2)y' + y^3 + y' = 3

Solve for y' in terms of x and y.

y'[3xy^2+1] = 3 - y^3
y' = [3 - y^3] / [3xy^2 + 1]

Answer 3

If you actually want dy/dz then the answer is 0.

Because y is not changing with respect to z, it is changing with respect to x.

3xydy/dx + y^3 +dy/dx = 3
(3xy + 1)dy/dx = 3 - y^3

dy/dx = (3 - y^3) / (3xy^2 + 1)

There's your answer.