Implicit Differentiation in multivariate calculus Let $y(x)$ be the be given explicitly by the equation : 
$xy\left( x\right) -\ln y\left( x\right) = 1$ 
Determine $\dfrac {dy}{dx}$
I'm unsure of how to go about this problem. 
 A: Deriving with respect to $x$ we have:
$$
\frac{d}{dx}(xy)-\frac{d}{dx}(\ln y)+\frac{d}{dx}(-1)=0
$$
$$
y+x\frac{dy}{dx}-\frac{1}{y}\frac{dy}{dx}=0
$$
$$
\frac{dy}{dx}\left(x-\frac{1}{y} \right)=-y
$$
$$
\frac{dy}{dx}=\frac{y^2}{1-xy}
$$
A: I'm going to abbreviate $y(x)$ as $y$ and $\frac{dy}{dx}$ as $y'$. I assume you mean to differentiate with respect to $x$. Please correct me if I'm wrong.
For $xy -\ln y = 1$, we differentiate both sides as $y+xy'-\frac{1}{y}y'=0.$
Then, we write $-y=y'(x-\frac{1}{y})$. 
Finally $$y'=\frac{-y}{x-\frac{1}{y}}=\frac{-y^2}{xy-1}.$$
If you want to write the derivative entirely in terms of $x$, you need to use the $W$ Lambert function, as you cannot solve $xy-\ln y=1$ for y using only "elementary functions" (some consider the Lambert function elementary, but I do not).
A: NOTE - the original question, before it was changed, was to find $\frac{dy}{dt}$, so that is what this answer is answering.  Even though $t$ is not a part of the original equation, finding the derivative of one variable with respect to another that isn't present is very commonly used in related rates problems.  Since $dt$ is just a value like any other, it can be divided just like any other, just as long as you do it to both sides.
To find the answer, just take a differential and divide by $dt$.  Usually, in Calculus, people try to find whole derivatives, but really, finding the differential first and then dividing is best.  
Let's start with your problem:
$$xy - \ln y = 1$$
Now, take the differential:
$$x\cdot dy + y\cdot dx - \frac{dy}{y} = 0$$
Now that you have the differential, divide both sides by $dt$:
$$x\frac{dy}{dt} + y\frac{dx}{dt} - \frac{1}{y}\frac{dy}{dt} = 0$$
Now, solve for $\frac{dy}{dt}$:
$$x\frac{dy}{dt} + y\frac{dx}{dt} - \frac{1}{y}\frac{dy}{dt} = 0$$
$$x\frac{dy}{dt} + y\frac{dx}{dt} = \frac{1}{y}\frac{dy}{dt} $$
$$y\frac{dx}{dt} = \frac{1}{y}\frac{dy}{dt} - x\frac{dy}{dt} $$
$$y\frac{dx}{dt} = (\frac{1}{y} - x)\frac{dy}{dt}$$
$$\frac{y}{\frac{1}{y} - x}\frac{dx}{dt} = \frac{dy}{dt}$$
A: I solved $y(x)$ not $y'(x)$, I'm sorry.
$$xy(x)-\ln(y(x))=1\Longleftrightarrow$$
$$\frac{\text{d}}{\text{d}x}\left[xy(x)-\ln(y(x))\right]=\frac{\text{d}}{\text{d}x}\left[1\right]\Longleftrightarrow$$
$$xy'(x)+y(x)-\frac{y'(x)}{y(x)}=0\Longleftrightarrow$$
$$y(x)+y'(x)\left[x-\frac{1}{y(x)}\right]=0\Longleftrightarrow$$

Let $\text{A}(x,y)=y$ and $\text{Q}(x,y)=x-\frac{1}{y}$,
This is an exact equation, because $\frac{\partial\text{A}(x,y)}{\partial y}=1=\frac{\partial\text{Q}(x,y)}{\partial x}$.
Define $f(x,y)$ such that $\frac{\partial f(x,y)}{\partial x}=\text{A}(x,y)$ and $\frac{\partial f(x,y)}{\partial y}=\text{Q}(x,y)$:
Then, the solution will be given by $f(x,y)=\text{C}$, where $\text{C}$ is an arbitrary constant.
Integrate $\frac{\partial f(x,y)}{\partial x}$ with respect to $x$ in order to find $f(x,y)$:
$f(x,y)=\int y\space\text{d}x=yx+g(y)$ where $g(y)$ is an arbitrary function of $y$.
Differentiate $f(x,y)$ with respect to $y$ in order to find $g(y)$:
$\frac{\partial f(x,y)}{\partial y}=\frac{\partial}{\partial y}\left[yx+g(y)\right]=x+\frac{\text{d}g(y)}{\text{d}y}$.
$$x+\frac{\text{d}g(y)}{\text{d}y}=x-\frac{1}{y}\Longleftrightarrow$$
$$\frac{\text{d}g(y)}{\text{d}y}=-\frac{1}{y}\Longleftrightarrow$$
$$g(y)=-\ln|y|\Longleftrightarrow$$
$$f(x,y)=y(x)-\ln|y|\Longleftrightarrow$$
$$f(x,y)=y(x)-\ln|y|$$

$$yx-\ln|y|=f(x,y)\Longleftrightarrow$$
$$yx-\ln|y|=\text{C}\Longleftrightarrow$$
$$y(x)=-\frac{\text{W}\left(\text{C}x\right)}{x}$$
