Answer
$$\boxed{y'=-\frac{2y(2xy+1)}{x(4xy+1}}$$
Work Step by Step
4xy+$\ln(x^{2}y)$=7
4xy+2$\ln$x+$\ln$y=7
Take $\frac{d}{dx}$ on both sides of the equation
4x(y')+4y(1)+$\frac{2}{x}$+$\frac{y'}{y}$=0
Make y' the subject of the equation
y'\left(4x+$\frac{1}{y}$\right)=-\left($\frac{2}{x}$+4y\right)
y'=-$\frac{\frac{2}{x}+4y}{4x+\frac{1}{y}}$
$\boxed{y'=-\frac{2y(2xy+1)}{x(4xy+1}}$
