Calculus question!!!!!

Find dy/dx by implicit differentiation and evaluate the derivative at the given point.

(4x + 4y)^3 = 64x^3 + 64y^3 (-1,1)



dy/dx = ???

At (−1, 1): y' = ???

(4x + 4y)^3 = 64x^3 + 64y^3
64x^3 + 192x²y + 192xy² + 64y^3 = 64x^3 + 64y^3
192x²y + 192xy² = 0
x²y + xy² = 0
2xy + x²y' + y² + 2xyy' = 0
x²y' + 2xyy' = -2xy - y²
y'(x² + 2xy) = -2xy - y²
y' = (-2xy - y²) / (x² + 2xy)
y'(-1,1) = [-2(-1)(1) - 1²] / [(-1)² + 2(-1)(1)]
y'(-1,1) = -1

So, dy/dx = (-2xy - y²) / (x² + 2xy), and at (-1,1), y' = -1. Hope that helps :)

y' = dy/dx:

(4x + 4y)^3 = 64x^3 + 64y^3 Differentiate:

3(4x + 4y)^2 * (4 + 4*y') = 192x^2 + 192y^2 * y'

12(4x + 4y)^2 + 12*y' * (4x + 4y)^2 = 192x^2 + 192y^2 * y'

(4x + 4y)^2 + y' * (4x + 4y)^2 = 16x^2 + 16y^2 * y'

y' * (4x + 4y)^2 - 16y^2 * y' = 16x^2 - (4x + 4y)^2

y' * [(4x + 4y)^2 - 16y^2] = 16x^2 - (4x + 4y)^2

y' = [16x^2 - (4x + 4y)^2] / [(4x + 4y)^2 - 16y^2]

At (- 1, 1):

[16*(- 1)^2 - (4(- 1) + 4(1))^2] / [(4(- 1) + 4(1))^2 - 16(1)^2]

[16 - (0)^2] / [(0)^2 - 16] = 16 / (- 16) = - 1