Derivative of arctan(1/x): Proof and the Minus Sign

Chain rule with inner function 1/x

The outer function is $\arctan u$, whose derivative is $\frac{1}{1+u^{2}}$. The inner function is $u = \frac{1}{x} = x^{-1}$, so $u' = -\frac{1}{x^{2}}$. Substituting $u^{2} = \frac{1}{x^{2}}$ produces a fraction inside a fraction.

The clean way through the algebra is to multiply the numerator and the denominator by $x^{2}$. Every term has to be multiplied, including the $1$: the denominator $1+\frac{1}{x^{2}}$ becomes $x^{2}+1$, and the numerator $-\frac{1}{x^{2}}$ becomes $-1$.

Why arctan(1/x) + arctan x is not one constant

The result $-\frac{1}{1+x^{2}}$ is exactly the negative of $\frac{d}{dx}\arctan x$. Add the two functions and the derivatives cancel, so $\arctan\left(\frac{1}{x}\right)+\arctan x$ has derivative zero wherever it is defined.

Zero derivative forces a constant only on an interval, and the domain here is missing $x=0$. That leaves two separate intervals, and the constant is different on each one. Testing $x=1$ gives $\frac{\pi}{4}+\frac{\pi}{4}$, while $x=-1$ gives $-\frac{\pi}{4}-\frac{\pi}{4}$.

The values $x = 0.5$, $1$ and $3$ all sit on the positive branch. The derivative formula itself makes no distinction between the branches, which is precisely why the jump in the antiderivative catches people out.

The mistakes students make

Two of these are algebra slips and the third is a piece of faulty reasoning about constants.

• Answering $\frac{1}{1+x^{2}}$, which is the derivative of $\arctan x$, not of $\arctan\left(\frac{1}{x}\right)$. The inner derivative $-\frac{1}{x^{2}}$ carries a minus sign that survives to the end.
• Clearing the compound fraction halfway: multiplying only the $\frac{1}{x^{2}}$ term by $x^{2}$ turns the denominator into $1+1=2$ and gives the false answer $-\frac{1}{2x^{2}}$.
• Concluding that $\arctan\left(\frac{1}{x}\right)+\arctan x$ equals $\frac{\pi}{2}$ for all $x$. On $x<0$ the sum is $-\frac{\pi}{2}$, because $x=0$ splits the domain into two intervals.