Find dy/dx (xy+1)^3=x-y^2+8

Problem

(x*y+1)3=x−y2+8

Solution

1. Differentiate both sides with respect to x using the chain rule on the left side and the power rule on the right side.

d(x*y+1)/d(x)=d(x−y2+8)/d(x)

2. Apply the chain rule to the left side, treating y as a function of x

3*(x*y+1)2⋅d(x*y+1)/d(x)=1−2*yd(y)/d(x)

3. Apply the product rule to the term x*y inside the derivative.

3*(x*y+1)2⋅(y+xd(y)/d(x))=1−2*yd(y)/d(x)

4. Distribute the term 3*(x*y+1)2 to both parts of the product rule result.

3*y*(x*y+1)2+3*x*(x*y+1)2d(y)/d(x)=1−2*yd(y)/d(x)

5. Group all terms containing d(y)/d(x) on one side of the equation and the remaining terms on the other side.

3*x*(x*y+1)2d(y)/d(x)+2*yd(y)/d(x)=1−3*y*(x*y+1)2

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

d(y)/d(x)*(3*x*(x*y+1)2+2*y)=1−3*y*(x*y+1)2

7. Solve for dy/dx by dividing both sides by the expression in the parentheses.

d(y)/d(x)=(1−3*y*(x*y+1)2)/(3*x*(x*y+1)2+2*y)

Final Answer

d(y)/d(x)=(1−3*y*(x*y+1)2)/(3*x*(x*y+1)2+2*y)

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