Find y' for y = arctan x + arccot x
Find $y'$ for $y = \arctan x + \operatorname{arccot} x$.
Explanation of Method
The same cancellation, this time on the whole real line.
Step 1: Recall the standard derivatives
\[ (\arctan x)' = \frac{1}{1+x^{2}}, \qquad (\operatorname{arccot} x)' = -\frac{1}{1+x^{2}}, \qquad x \in \mathbb{R}. \]
Step 2: Differentiate term by term
\[ y' = \frac{1}{1+x^{2}} + \left(-\frac{1}{1+x^{2}}\right) = 0. \]
Step 3: Verification
Since $\mathbb{R}$ is an interval, a vanishing derivative forces $y$ to be constant everywhere; evaluating at $x = 0$ and $x = 1$ identifies the constant: \[x = 0:\quad y = 0 + \frac{\pi}{2} = \frac{\pi}{2}\]
\[x = 1:\quad y = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2} \;\;\checkmark\]
Hence the identity
\[ \arctan x + \operatorname{arccot} x = \frac{\pi}{2}, \qquad x \in \mathbb{R}. \]
Result
Final answer:
\[\boxed{\,y' = 0, \qquad x \in \mathbb{R}\,}\]
Remark
Remark. Unlike Problem 54, here the denominator $1+x^{2}$ never vanishes, so the argument is valid on all of $\mathbb{R}$ with no endpoints to worry about. Problem 58 puts this identity to work.