Find y' for x ln y - y ln x = 1, and evaluate y' at x = 1

Find $y'$ for $x\ln y - y\ln x = 1$, and evaluate $y'$ at $x = 1$.

Step 1: Differentiate term by term

Each term is a product, and in each one the second factor needs the chain rule. For the first term, $\bigl(\ln y\bigr)' = \dfrac{y'}{y}$: $\frac{d}{dx}\bigl(x\ln y\bigr) = \ln y + x\cdot\frac{y'}{y}$.

For the second term, $\bigl(\ln x\bigr)' = \dfrac1x$: $\frac{d}{dx}\bigl(y\ln x\bigr) = y'\ln x + \frac{y}{x}$.

Step 2: Set the derivative of the relation to zero

The right-hand side is the constant $1$, so the whole derivative vanishes: $\ln y + \frac{xy'}{y} - y'\ln x - \frac{y}{x} = 0$.

Step 3: Collect the $y'$ terms

\[ y'\left(\frac{x}{y} - \ln x\right) = \frac{y}{x} - \ln y. \]

Step 4: Clear the fractions

Multiplying numerator and denominator by $xy$ removes both inner fractions at once: \[y' = \frac{\dfrac{y}{x}-\ln y}{\dfrac{x}{y}-\ln x} = \frac{xy\left(\dfrac{y}{x}-\ln y\right)}{xy\left(\dfrac{x}{y}-\ln x\right)} = \frac{y^{2}-xy\ln y}{x^{2}-xy\ln x}.\]

Step 5: Evaluate at $x = 1$

First locate the point. Putting $x = 1$ in the original relation makes $\ln x = 0$: \[1\cdot\ln y - y\cdot 0 = 1 \;\Longrightarrow\; \ln y = 1 \;\Longrightarrow\; y = e.\]

Now substitute $x = 1,\ y = e,\ \ln y = 1,\ \ln x = 0$ into the answer: \[y'(1) = \frac{e^{2}-1\cdot e\cdot 1}{1^{2}-1\cdot e\cdot 0} = \frac{e^{2}-e}{1} = e^{2}-e \approx 4.67077.\]

Step 6: Verification

\[\text{Point check: } x=1,\ y=e:\quad 1\cdot\ln e - e\cdot\ln 1 = 1-0 = 1 \;\;\checkmark\]

\[x = 0.999 \;\Rightarrow\; y \approx 2.713619, \qquad x = 1.001 \;\Rightarrow\; y \approx 2.722961\]

\[\frac{2.722961-2.713619}{0.002} \approx 4.6708 \approx e^{2}-e \;\;\checkmark\]