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


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

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*}

Compute the derivatives of two given functions

Consider the functions

    \begin{align*}  f(x) &= \frac{1}{1 + \frac{1}{x}}, \qquad x \neq 0 \\  g(x) &= \frac{1}{1 + \frac{1}{f(x)}}. \end{align*}

Find thd derivatives f'(x) and g'(x).

First, we compute the derivative of f,

    \begin{align*}  f'(x) &= \frac{\frac{1}{x^2}}{\left( 1+ \frac{1}{x} \right)^2} \\  &= \frac{1}{(x+1)^2}, \end{align*}

where we multiplied the numerator and denominator by x^2 to get the second line.

Next, using the chain rule we know

    \[ \left( 1+ \frac{1}{f(x)} \right)' = \frac{-f'(x)}{(f(x))^2}. \]

So, then using the quotient rule and chain rule we can evaluate the derivative of g,

    \begin{align*}  g'(x) &= \frac{\frac{f'(x)}{(f(x))^2}}{\left(1+ \frac{1}{f(x)}\right)^2} \\  &= \frac{f'(x)}{(f(x))^2 \left( 1 + \frac{1}{f(x)} \right)^2} \\  &= \frac{f'(x)}{(f(x) + 1)^2}. \end{align*}

Then, plugging in the definition of f(x) and the expression for f'(x) which we already computed, we have,

    \begin{align*}  g'(x) &= \frac{ \frac{1}{(x+1)^2} }{\left( \frac{1}{1+ \frac{1}{x}} + 1 \right)^2} \\[9pt]  &= \frac{1}{(1+x)^2 \left( \frac{2x+1}{x+1} \right)^2} \\  &= \frac{1}{(2x+1)^2} \\  &= (2x+1)^{-2}.  \end{align*}