1. ## find dy/dx

• find $\displaystyle \frac{dy}{dx}$ if $\displaystyle ln(xy)=x+y$

this is what l tried to do to solve it:
1. $\displaystyle ln(xy)=x+y$
2. $\displaystyle lnx+lny=x+y$
3. $\displaystyle y=lnx+lny-x$
4. $\displaystyle y'=\frac{1}{x}+\frac{1}{y}-1$
5. $\displaystyle y'=\frac{y+x}{xy}-1$
6. $\displaystyle y'=\frac{y+x-xy}{xy}$

however, the answer is suppose to be $\displaystyle \frac{xy-y}{x-xy}$

how do you solve this?

2. Originally Posted by algebra2
3. $\displaystyle y=lnx+lny-x$
4. $\displaystyle y'=\frac{1}{x}+\color{red}\frac{1}{y}\color{black}-1$
No. Remember that you are differentiating with respect to $\displaystyle x,$ so you must use the chain rule: $\displaystyle \frac d{dx}[\ln y]=\frac1y\frac{dy}{dx}.$

For this problem, just use implicit differentiation, as follows.

$\displaystyle \ln xy=x+y$

$\displaystyle \Rightarrow\frac d{dx}\left[\ln xy\right]=\frac d{dx}\left[x+y\right]$

$\displaystyle \Rightarrow\frac d{dx}\left[\ln x+\ln y\right]=1+\frac{dy}{dx}$

$\displaystyle \Rightarrow\frac1x+\frac{dy/dx}y=1+\frac{dy}{dx}$

$\displaystyle \Rightarrow\frac{dy/dx}y-\frac{dy}{dx}=1-\frac1x$

$\displaystyle \Rightarrow\frac{dy}{dx}\left(\frac1y-1\right)=\frac{x-1}x$

$\displaystyle \Rightarrow\frac{dy}{dx}=\frac{x-1}x\cdot\frac y{1-y}$

$\displaystyle \Rightarrow\frac{dy}{dx}=\frac{xy-y}{x-xy}$