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Find the derivative of arctan ((1+x)/(1-x))

by RoRi

Find the derivative of the function

    \[ f(x) = \arctan \frac{1+x}{1-x}. \]

Using the chain rule and the formula for the derivative of arctangent we have

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

This is valid for x \neq 1.

2 comments

    • RoRi says:

      The formula for the derivative of \arctan x is valid for all real x, but in this case we are looking at \arctan \left(\frac{x+1}{x-1}\right) and the expression \frac{x+1}{x-1} is not defined for x = 1, so our formula is only good when x \neq 1.

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