Find y' for y = arctan (1+x)/(1-x)

Find $y'$ for $y = \arctan\dfrac{1+x}{1-x}$.

Step 1: Apply the chain rule

With $u = \dfrac{1+x}{1-x}$: $y' = \frac{u'}{1+u^{2}}$.

Step 2: Differentiate the inner quotient

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

Step 3: Compute $1+u^{2}$

The cross terms cancel when the two squares are added: \[1+u^{2} = \frac{(1-x)^{2}+(1+x)^{2}}{(1-x)^{2}} = \frac{2+2x^{2}}{(1-x)^{2}} = \frac{2\bigl(1+x^{2}\bigr)}{(1-x)^{2}}.\]

Step 4: Combine

The factors $(1-x)^{2}$ and $2$ both cancel: $y' = \frac{(1-x)^{2}}{2\bigl(1+x^{2}\bigr)}\cdot\frac{2}{(1-x)^{2}} = \frac{1}{1+x^{2}}$.

Step 5: Verification

The result says $y$ differs from $\arctan x$ by a constant on each interval of its domain. Checking the constant at a point on each side of $x = 1$: \[x = 0:\quad y = \arctan 1 = \frac{\pi}{4},\quad \arctan 0 = 0 \;\Longrightarrow\; y - \arctan x = \frac{\pi}{4}\]

\[x = 2:\quad y = \arctan(-3) \approx -1.24905,\quad \arctan 2 \approx 1.10715 \;\Longrightarrow\; y - \arctan x \approx -2.35619 = \frac{\pi}{4}-\pi\]

Step 6: A caution about the identity

It is tempting to write $\arctan\dfrac{1+x}{1-x} = \dfrac{\pi}{4} + \arctan x$, and on $x \lt 1$ that is correct. For $x \gt 1$ the addition formula for the arctangent picks up a shift of $\pi$, and the true identity is

\[ \arctan\frac{1+x}{1-x} = \begin{cases} \dfrac{\pi}{4} + \arctan x, & x \lt 1, \\[8pt] \dfrac{\pi}{4} + \arctan x - \pi, & x \gt 1. \end{cases} \]

The derivative is the same on both branches, which is why the computation above is valid throughout — the constant differs, but constants differentiate to zero.