Find dy/dx (4x+4y)^3=64x^3+64y^3

Problem

(4*x+4*y)3=64*x3+64*y3

Solution

1. Simplify the equation by factoring out the constant 4 from the left side.

(4*(x+y))3=64*x3+64*y3

64*(x+y)3=64*x3+64*y3

2. Divide both sides by 64 to isolate the binomial cube.

(x+y)3=x3+y3

3. Expand the left side using the binomial expansion formula (a+b)3=a3+3*a2*b+3*a*b2+b3

x3+3*x2*y+3*x*y2+y3=x3+y3

4. Subtract x3 and y3 from both sides to simplify the equation.

3*x2*y+3*x*y2=0

5. Factor the resulting expression.

3*x*y*(x+y)=0

6. Differentiate implicitly with respect to x Since the product equals zero, we can differentiate the simplified form 3*x2*y+3*x*y2=0 using the product rule.

(d(3)*x2*y)/d(x)+(d(3)*x*y2)/d(x)=0

(6*x*y+3*x2d(y)/d(x))+(3*y2+6*x*yd(y)/d(x))=0

7. Group the terms containing d(y)/d(x) on one side.

(3*x2+6*x*y)d(y)/d(x)=−6*x*y−3*y2

8. Solve for d(y)/d(x) by dividing.

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

d(y)/d(x)=−(y*(2*x+y))/(x*(x+2*y))

Final Answer

d(y)/d(x)=−(2*x*y+y2)/(x2+2*x*y)

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