Find y' for y = arctan x - 1/x
Find $y'$ for $y = \arctan x - \dfrac{1}{x}$.
Explanation of Method
A sum of two standard derivatives over a common denominator.
Step 1: Differentiate term by term
Note that subtracting $\dfrac1x$, whose derivative is $-\dfrac{1}{x^{2}}$, contributes a positive term: \[\frac{d}{dx}\bigl(\arctan x\bigr) = \frac{1}{1+x^{2}}, \qquad \frac{d}{dx}\!\left(-\frac{1}{x}\right) = \frac{1}{x^{2}}.\]
Step 2: Combine over a common denominator
\[ y' = \frac{1}{1+x^{2}} + \frac{1}{x^{2}} = \frac{x^{2} + \bigl(1+x^{2}\bigr)}{x^{2}\bigl(1+x^{2}\bigr)} = \frac{2x^{2}+1}{x^{2}\bigl(1+x^{2}\bigr)}. \]
Step 3: Verification
\[x = 1:\quad y' = \frac{3}{1\cdot 2} = \frac32; \qquad \frac{1}{2}+\frac{1}{1} = \frac32 \;\;\checkmark\]
\[\frac{y(1+10^{-5})-y(1-10^{-5})}{2\times 10^{-5}} \approx 1.5 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = \dfrac{2x^{2}+1}{x^{2}\bigl(1+x^{2}\bigr)}, \qquad x \neq 0\,}\]
Remark
Remark. Both terms of the derivative are positive, so $y$ increases on each of $(-\infty,0)$ and $(0,\infty)$ separately — but not across the origin, where $-\tfrac1x$ has a pole. Compare Problem 59, where the same two functions appear with the rational term built differently and the derivative comes out as a single monomial fraction.