Find y' for y = (arctan x + arccot x)^n, where n is

Find $y'$ for $y = \bigl(\arctan x + \operatorname{arccot} x\bigr)^{n}$, where $n$ is a constant.

Step 1: Apply the identity from Problem 55

The base of the power is not a genuine function of $x$ at all: $\arctan x + \operatorname{arccot} x = \frac{\pi}{2}, \qquad x \in \mathbb{R}$.

Therefore $y = \left(\frac{\pi}{2}\right)^{\!n}$.

Step 2: Differentiate a constant

The expression $\left(\dfrac{\pi}{2}\right)^{n}$ contains no $x$, so it is a constant and

\[ y' = 0. \]

Step 3: Verification

The chain rule reaches the same conclusion the long way round, using $y' = n u^{n-1} u'$ with $u = \arctan x + \operatorname{arccot} x$: \[u' = \frac{1}{1+x^{2}} - \frac{1}{1+x^{2}} = 0 \;\Longrightarrow\; y' = n\left(\frac{\pi}{2}\right)^{n-1}\cdot 0 = 0 \;\;\checkmark\]