Using the derivative of an inverse function, prove that

Using the derivative of an inverse function, prove that $\left(\arctan\dfrac{1}{x}\right)' = -\dfrac{1}{1+x^{2}}$.

Step 1: Set up the inverse relation

Let $y = \arctan\dfrac{1}{x}$ with $x \neq 0$. Then $\tan y = \dfrac1x$, and taking reciprocals, $x = \frac{1}{\tan y} = \cot y$.

So $x$ is expressed explicitly as a function of $y$ — exactly what the inverse rule needs.

Step 2: Differentiate the inverse

\[ x'_{y} = \bigl(\cot y\bigr)' = -\frac{1}{\sin^{2}y}. \]

Step 3: Apply the inverse function rule

\[ y'_{x} = \frac{1}{x'_{y}} = \frac{1}{-\dfrac{1}{\sin^{2}y}} = -\sin^{2}y. \]

Step 4: Express in terms of $x$

From $\tan y = \dfrac1x$ and the identity $\sin^{2}y = \dfrac{\tan^{2}y}{1+\tan^{2}y}$: \[\sin^{2}y = \frac{\dfrac{1}{x^{2}}}{1+\dfrac{1}{x^{2}}} = \frac{1}{x^{2}+1}, \qquad\text{hence}\qquad y'_{x} = -\frac{1}{1+x^{2}}.\]

Step 5: Verification by the chain rule

The direct route gives the same answer: \[\left(\arctan\frac1x\right)' = \frac{1}{1+\dfrac{1}{x^{2}}}\cdot\left(-\frac{1}{x^{2}}\right) = \frac{x^{2}}{x^{2}+1}\cdot\left(-\frac{1}{x^{2}}\right) = -\frac{1}{1+x^{2}} \;\;\checkmark\]

\[x = 1:\quad y' = -\tfrac12; \qquad \frac{\arctan\frac{1}{1.001}-\arctan\frac{1}{0.999}}{0.002} \approx -0.5 \;\;\checkmark\]