Answer
$\frac{dy}{dx}=\frac{8x^3\sqrt{x+y}-1}{1-8y^3\sqrt{x+y}}$
Work Step by Step
$\sqrt {x+y} = x^4 + y^4$
Differentiate both sides with respect to x.
$\frac{1+\frac{dy}{dx}}{2\sqrt{x+y}} = 4x^3 + 4y^3\frac{dy}{dx}$
Isolate $\frac{dy}{dx}$
$1+\frac{dy}{dx} = 2\sqrt{x+y}(4x^3 + 4y^3\frac{dy}{dx})$
$\frac{dy}{dx} - 8y^3\sqrt{x+y}\frac{dy}{dx} = 8x^3\sqrt{x+y}-1$
$\frac{dy}{dx}(1-8y^3\sqrt{x+y}) = 8x^3\sqrt{x+y}-1$
$\frac{dy}{dx}=\frac{8x^3\sqrt{x+y}-1}{1-8y^3\sqrt{x+y}}$
