Find dy/dx 2x^3+x^2y-xy^3=2

Problem

2*x3+x2*y−x*y3=2

Solution

1. Differentiate both sides of the equation with respect to x treating y as a function of x and applying the chain rule where necessary.

(d(2)*x3)/d(x)+(d(x2)*y)/d(x)−(d(x)*y3)/d(x)=d(2)/d(x)

2. Apply the product rule to the terms x2*y and x*y3

6*x2+(x2d(y)/d(x)+2*x*y)−(x⋅3*y2d(y)/d(x)+y3)=0

3. Distribute the negative sign and group all terms containing d(y)/d(x) on one side of the equation.

6*x2+x2d(y)/d(x)+2*x*y−3*x*y2d(y)/d(x)−y3=0

x2d(y)/d(x)−3*x*y2d(y)/d(x)=y3−2*x*y−6*x2

4. Factor out d(y)/d(x) from the left side.

d(y)/d(x)*(x2−3*x*y2)=y3−2*x*y−6*x2

5. Solve for d(y)/d(x) by dividing both sides by the expression in the parentheses.

d(y)/d(x)=(y3−2*x*y−6*x2)/(x2−3*x*y2)

Final Answer

d(y)/d(x)=(y3−2*x*y−6*x2)/(x2−3*x*y2)

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