# implicit differentation

1.y^3=4(x^2+y^2)

2.y^2-3x+2y=0

help please

You consider y to be a function of x, but you don't explicitely solve for it. Then you formally differentiate w.r.t. x using the chain rule:

y^3(x) = 4(x^2+y^2(x)) --->

3y^2dy/dx = 8x + 8y dy/dx

y^2-3x+2y=0 --->

2ydy/dx - 3+2dy/dx = 0

## To solve these problems using implicit differentiation, you need to treat y as a function of x. Here's how you can solve each problem:

1. y^3 = 4(x^2 + y^2):

Step 1: Differentiate both sides of the equation with respect to x. Remember to use the chain rule when differentiating y^2(x).

d/dx(y^3) = d/dx(4(x^2 + y^2))

Step 2: Apply the power rule for differentiation to y^3. This states that d/dx(y^n) = ny^(n-1) * dy/dx.

3y^2 * (dy/dx) = 8x + 8y * (dy/dx)

Step 3: Rearrange the equation and solve for dy/dx.

3y^2 * (dy/dx) - 8y * (dy/dx) = 8x

(dy/dx) * (3y^2 - 8y) = 8x

dy/dx = (8x) / (3y^2 - 8y)

2. y^2 - 3x + 2y = 0:

Step 1: Differentiate both sides of the equation with respect to x. Treat y as a function of x and use the chain rule when differentiating.

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

Step 2: Apply the power rule for differentiation to y^2 and differentiate the other terms.

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

Step 3: Rearrange the equation and solve for dy/dx.

2y * (dy/dx) + 2 * (dy/dx) = 3

(dy/dx) * (2y + 2) = 3

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

These are the implicit differentiation solutions for the given equations. Remember to simplify your answers if possible.