Find y' for arctan y/x - ln sqrt(x²+y²) = 0, and evaluate

Find $y'$ for $\arctan\dfrac{y}{x} - \ln\sqrt{x^{2}+y^{2}} = 0$, and evaluate it where $y = 0$.

Step 1: Find $x$ where $y = 0$

Setting $y = 0$ gives $\arctan 0 = 0$, so the equation reduces to

\[ \ln\sqrt{x^{2}} = \ln|x| = 0 \;\Longrightarrow\; |x| = 1. \]

Taking the positive branch, the point is $(1,0)$.

Step 2: Differentiate the arctangent

With $u = \dfrac{y}{x}$, the quotient rule gives $u' = \dfrac{y'x-y}{x^{2}}$, and the $x^{2}$ cancels: \[\frac{d}{dx}\arctan\frac{y}{x} = \frac{1}{1+\dfrac{y^{2}}{x^{2}}}\cdot\frac{y'x-y}{x^{2}} = \frac{x^{2}}{x^{2}+y^{2}}\cdot\frac{y'x-y}{x^{2}} = \frac{y'x-y}{x^{2}+y^{2}}.\]

Step 3: Differentiate the logarithm

Pull the root out first: $\ln\sqrt{x^{2}+y^{2}} = \tfrac12\ln\bigl(x^{2}+y^{2}\bigr)$, so

\[ \frac{d}{dx}\!\left[\frac12\ln\bigl(x^{2}+y^{2}\bigr)\right] = \frac12\cdot\frac{2x+2y\,y'}{x^{2}+y^{2}} = \frac{x+y\,y'}{x^{2}+y^{2}}. \]

Step 4: Equate and solve

Both expressions share the denominator $x^{2}+y^{2}$, which cancels immediately: \[\frac{y'x-y}{x^{2}+y^{2}} - \frac{x+y\,y'}{x^{2}+y^{2}} = 0 \;\Longrightarrow\; y'x - y - x - y\,y' = 0,\]

\[ y'\bigl(x-y\bigr) = x+y \;\Longrightarrow\; y' = \frac{x+y}{x-y}. \]

Step 5: Evaluate at $(1,0)$

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

Step 6: What the curve is

In polar coordinates $x = \rho\cos\theta$, $y = \rho\sin\theta$, the equation reads $\theta = \ln\rho$, that is

\[\rho = e^{\theta} \text{ — the logarithmic spiral of Problem 197}\]

\[\theta = 0:\ \rho = 1 \;\Longrightarrow\; \text{the point } (1,0), \text{ with slope } 1 \text{, i.e. } 45^{\circ} \;\;\checkmark\]