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Prove a property of the derivative if arctangent and the logarithm obey a given relation

If

    \[ \arctan \frac{y}{x} = \log \sqrt{x^2 + y^2} \]

prove that

    \[ \frac{dy}{dx} = \frac{x+y}{x-y}. \]


Proof. First, we consider the derivatives of the left and right side of the given equation. (Treating y as a function of x and remembering to use the chain rule.) So, for the derivative on the left, we have

    \begin{align*}  D \left( \arctan \frac{y}{x} \right) &= \left( \frac{-y}{x^2} + \frac{1}{x} \frac{dy}{dx} \right) \left( \frac{1}{1+\left( \frac{y}{x} \right)^2} \right) \\ &= \frac{x \frac{dy}{dx} - y}{x^2+y^2}. \end{align*}

On the right we have,

    \begin{align*}  D \left( \log \sqrt{x^2+y^2} \right) &= \left( x + y \frac{dy}{dx} \right) \left( \frac{1}{x^2+y^2} \right)\\  &= \frac{x + y \frac{dy}{dx}}{x^2+y^2}. \end{align*}

Now, using the given equation we have

    \begin{align*}  \arctan \frac{y}{x} = \log \sqrt{x^2+y^2} && \implies && D \left( \arctan \frac{y}{x} \right) &= D \left( \log \sqrt{x^2+y^2} \right) \\[9pt]  && \implies && \frac{x \frac{dy}{dx} - y}{x^2+y^2} &= \frac{x+y\frac{dy}{dx}}{x^2+y^2} \\[9pt]  && \implies && x \frac{dy}{dx} - y &= x + y\frac{dy}{dx} \\  && \implies && (x-y) \frac{dy}{dx} &= x + y \\  && \implies && \frac{dy}{dx} &= \frac{x+y}{x-y}. \qquad \blacksquare \end{align*}

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