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

Problem

x3+3*x2*y+y3=8

Solution

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

d(x3)/d(x)+(d(3)*x2*y)/d(x)+d(y3)/d(x)=d(8)/d(x)

2. Apply the product rule to the term 3*x2*y and the power rule to the other terms.

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

3. Group the terms containing d(y)/d(x) on one side and move the remaining terms to the other side.

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

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

d(y)/d(x)*(3*x2+3*y2)=−3*x2−6*x*y

5. Solve for d(y)/d(x) by dividing both sides by (3*x2+3*y2)

d(y)/d(x)=(−3*x2−6*x*y)/(3*x2+3*y2)

6. Simplify the fraction by dividing the numerator and denominator by 3.

d(y)/d(x)=(−(x2+2*x*y))/(x2+y2)

Final Answer

d(y)/d(x)=−(x2+2*x*y)/(x2+y2)

Want more problems? Check here!