4. **find dy/dx when x = 1 of ln(x - y) = x.

Explanation:

Step1: Differentiate both sides

Differentiate $\ln(x - y)$ and $x$ with respect to $x$. The derivative of $\ln(u)$ with respect to $x$ is $\frac{u'}{u}$ (chain - rule), and the derivative of $x$ with respect to $x$ is 1. So, $\frac{1 - \frac{dy}{dx}}{x - y}=1$.

Step2: Solve for $\frac{dy}{dx}$

Multiply both sides by $x - y$: $1-\frac{dy}{dx}=x - y$. Then, isolate $\frac{dy}{dx}$: $\frac{dy}{dx}=1-(x - y)=1 - x + y$.

Step3: Find $y$ when $x = 1$

When $x = 1$, $\ln(1 - y)=1$. Then, $1 - y=e$, so $y = 1 - e$.

Step4: Substitute $x = 1$ and $y = 1 - e$ into $\frac{dy}{dx}$

$\frac{dy}{dx}\big|_{x = 1}=1-1+(1 - e)=1 - e$.

Answer:

$1 - e$