<s>[INST] Find $\frac{dy}{dx}$ implicitly given the equation $5xy+\ln(xy^2)=2$ Find $\frac{dy}{dx}$ implicitly given the equation $5xy+\ln(xy^2)=2$
edit: the homework might be $5xy+\ln(x^2y)=2$, but this is close enough to get the point

$Solution:$
First lets use properties of $\ln$ to write our equation in a way that will be easier to take the derivative.
$5xy+\ln(xy^2)=2$
$\rightarrow 5xy+\ln(x)+\ln(y^2)=2$
$\rightarrow 5xy+\ln(x)+2\ln(y)=2$
Okay, now lets take our implicit derivative, making sure to use the product rule on $5xy$:
$(5xy+\ln(x)+2\ln(y))'=2'=0$
$\rightarrow 5(xy)'+\ln(x)'+2\ln(y)'=0$
$\rightarrow 5(x'y+xy')+\frac{1}{x}+2\frac{1}{y}y'=0$
recall that $x' = \frac{d}{dx}x=1$, while $y' = \frac{d}{dx}y=\frac{dy}{dx}$
$\rightarrow 5((1)y+xy')+\frac{1}{x}+2\frac{1}{y}y'=0$
$\rightarrow 5y+5xy'+\frac{1}{x}+2\frac{1}{y}y'=0$
multiplying both sides of the equation by $x$ yields:
$\rightarrow 5xy+5x^2y'+ 1 + 2\frac{x}{y}y'=0(x)=0$
now we multiply both sides of the equation by $y$:
$\rightarrow 5xy^2+5x^2yy'+y+2xy'=0(y)=0$
$\rightarrow 5x^2yy'+2xy'=-5xy^2-y$
$\rightarrow (5x^2y+2x)y'=-5xy^2-y$
$\rightarrow y'=\frac{-5xy^2-y}{5x^2y+2x}$
 [/INST] $$5xy+ \ln(xy^2)=2$$
Differentiating w.r.t $x$
$$5y+5xy'+\frac{y^2+2xyy'}{xy^2}=0$$
$$y'=-\frac{(5xy+1)y}{x(2+5xy)}=\frac{dy}{dx}$$
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