find dy/dx (implicit differentiation) 3xy^2+y^3=8
pls do step by step
To find the derivative of the given equation implicitly with respect to x, follow these steps:
1. Differentiate both sides of the equation with respect to x:
d/dx (3xy^2) + d/dx (y^3) = d/dx (8)
2. Apply the product rule to the first term:
d/dx(3xy^2) = 3y^2 * d/dx(x) + 3x * d/dx(y^2)
The product rule states that if you have a function f(x) = g(x) * h(x), then the derivative, f'(x) = g'(x) * h(x) + g(x) * h'(x).
3. Apply the chain rule to d/dx(y^2):
d/dx(y^2) = 2y * dy/dx
The chain rule states that if you have a function y(x), then the derivative of y^n is n * y^(n-1) * dy/dx.
4. Continue differentiating the other terms:
d/dx(y^3) = 3y^2 * dy/dx
d/dx(8) = 0
Now put all the pieces together:
3y^2 * d/dx(x) + 3x * (2y * dy/dx) + 3y^2 * dy/dx = 0
5. Simplify the equation:
3y^2 + 6xy * dy/dx + 3y^2 * dy/dx = 0
6. Solve for dy/dx:
dy/dx * (6xy + 3y^2) = -3y^2
dy/dx = (-3y^2) / (6xy + 3y^2)
7. Simplify the expression:
dy/dx = -y^2 / (2x + y^2)
The derivative dy/dx is:
dy/dx = -y^2 / (2x + y^2)