Question

How do you find dydx given 4x2+4xy=−5x3y+4?

Answer 1

8+4xy=20×3y

8+20=4xy×3y

28=12xy²

Answer 2

$Given \\ 4x^2+4xy = -5x^3y+4 \\ \frac{d}{dx}(4x^2+4xy) = \frac{d}{dx}(-5x^3y+4) \\ 4(2x)+4y+4x\frac{dy}{dx}=-5(3x^2y)\ -\ 5x^3\frac{dy}{dx}+0 \\ 8x+4y+4x\frac{dy}{dx}=-15x^2y\ -\ 5x^3\frac{dy}{dx} \\ 4x\frac{dy}{dx}+5x^3\frac{dy}{dx}=-15x^2y\ -\ 8x\ -\ 4y \\ (4x+5x^3)\frac{dy}{dx}=-15x^2y\ -\ 8x\ -\ 4y \\ \bold{\frac{dy}{dx}=\frac{-15x^2y\ -\ 8x\ -\ 4y}{4x+5x^3}}$
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