How do you differentiate #y^3-y=x^3-y+3xy#?

Question

Samantha Adkison · Accepted Answer

Please see the explanation section below.

Begin by rewriting. Add #y# to both sides to get rid of the duplicate terms. #y^3=x^3+3xy#. Note that the term #3xy# has a product of both variables in it. When we differentiate that term we will need the product rule. Now, assuming that you are differentiating with respect to #x#, we get #d/dx(y^3) = d/dx(x^3) + d/dx(3xy)#. So #3y^2 dy/dx = 3x^2 + [3y+3x dy/dx]# Solving for #dy/dx#, we have #3y^2 dy/dx - 3x dy/dx = 3x^2 +3y# #y^2 dy/dx - x dy/dx = x^2 +y# #(y^2 - x) dy/dx = x^2 +y# So, #dy/dx = (x^2 +y)/(y^2 - x)# If you are differentiating with respect to #t# we get: #3y^2 dy/dt = 3x^2 dx/dt +3y dx/dt +3x dy/dt#. So, #(y^2-x) dy/dx = (x^2+y) dx/dt#.

Charles Anctil · Answer

undefined To differentiate the equation $y^3 - y = x^3 - y + 3xy$, follow these steps:

1. Differentiate both sides of the equation with respect to $x$.
2. Use the chain rule and product rule as necessary.
3. Solve for $\frac{{dy}}{{dx}}$.

Here are the steps in detail:

$$ y^3 - y = x^3 - y + 3xy $$

1. Differentiate both sides with respect to $x$:
$$ \frac{{d}}{{dx}}(y^3 - y) = \frac{{d}}{{dx}}(x^3 - y + 3xy) $$

2. Apply the chain rule and power rule to differentiate $y^3$ with respect to $x$:
$$ 3y^2 \frac{{dy}}{{dx}} - \frac{{dy}}{{dx}} = 3x^2 - \frac{{dy}}{{dx}} + 3y + 3x\frac{{dy}}{{dx}} $$

3. Rearrange terms to solve for $\frac{{dy}}{{dx}}$:
$$ 3y^2 \frac{{dy}}{{dx}} - \frac{{dy}}{{dx}} + \frac{{dy}}{{dx}} - 3x\frac{{dy}}{{dx}} = 3x^2 + 3y $$

$$ 3y^2 \frac{{dy}}{{dx}} - 3x\frac{{dy}}{{dx}} = 3x^2 + 3y $$

$$ \frac{{dy}}{{dx}}(3y^2 - 3x) = 3x^2 + 3y $$

$$ \frac{{dy}}{{dx}} = \frac{{3x^2 + 3y}}{{3y^2 - 3x}} $$

Thus, the derivative of $y$ with respect to $x$ is:

$$ \frac{{dy}}{{dx}} = \frac{{3x^2 + 3y}}{{3y^2 - 3x}} $$