## Calculus: Early Transcendentals 8th Edition

$y' = \frac{1-8x^{3}\sqrt {x+y}}{8y^{3}\sqrt {x+y} -1}$
Equation: $\sqrt{x+y} = x^{4} + y^{4}$. Differentiate both sides: $\frac{1+y'}{2\sqrt {x+y}} = 4x^{3} + 4y'y^{3}$. Simplify and solve for y': $1+y' = 8y'y^{3}\sqrt {x+y} + 8x^{3}\sqrt {x+y}$. $1-8x^{3}\sqrt {x+y} = y'(8y^{3}\sqrt {x+y} - 1)$. $y' = \frac{1-8x^{3}\sqrt {x+y}}{8y^{3}\sqrt {x+y} -1}$