Find the slope k of the tangent to the curve ln

Find the slope $k$ of the tangent to the curve $\ln\dfrac{y}{x}+x-y^{2}=0$ at the point $(1,1)$.

Step 1: Check that the point lies on the curve

\[ \ln\frac{1}{1}+1-1^{2} = 0+1-1 = 0 \;\;\checkmark \]

Step 2: Split the logarithm first

Before differentiating, use $\ln\dfrac{y}{x} = \ln y - \ln x$. This replaces a quotient inside a logarithm by two independent terms: $\ln y - \ln x + x - y^{2} = 0$.

Step 3: Differentiate with respect to $x$

\[ \frac{y'}{y} - \frac1x + 1 - 2yy' = 0. \]

Step 4: Solve for $y'$

\[ y'\left(\frac1y - 2y\right) = \frac1x - 1 \;\Longrightarrow\; y'\cdot\frac{1-2y^{2}}{y} = \frac{1-x}{x}, \]

\[ y' = \frac{y(1-x)}{x\bigl(1-2y^{2}\bigr)} = \frac{y(x-1)}{x\bigl(2y^{2}-1\bigr)}. \]

Step 5: Evaluate at $(1,1)$

The numerator contains the factor $x-1$, which vanishes at $x=1$; the denominator is $1\cdot(2-1)=1\neq 0$, so there is no indeterminacy: \[k = y'(1) = \frac{1\cdot(1-1)}{1\cdot(2\cdot 1-1)} = \frac{0}{1} = 0.\]

Step 6: Verification

\[x = 0.99 \;\Rightarrow\; y \approx 1.0000503, \qquad x = 1.01 \;\Rightarrow\; y \approx 1.0000497\]

\[\text{Both neighbours lie above the point, so the slope is } 0 \text{ and } (1,1) \text{ is a local minimum} \;\;\checkmark\]

\[y'' = \frac{y\Bigl[\bigl(2y^{2}-1\bigr)^{2}-(x-1)^{2}\bigl(2y^{2}+1\bigr)\Bigr]}{x^{2}\bigl(2y^{2}-1\bigr)^{3}}, \qquad y''(1) = \frac{1\cdot 1}{1\cdot 1} = 1 > 0 \;\;\checkmark\]