Find y' for y = arctan x - x/(1+x²)
Find $y'$ for $y = \arctan x - \dfrac{x}{1+x^{2}}$.
Explanation of Method
A difference whose derivative simplifies to a single monomial fraction.
Step 1: Differentiate the arctangent
\[ \frac{d}{dx}\bigl(\arctan x\bigr) = \frac{1}{1+x^{2}}. \]
Step 2: Differentiate the rational term by the quotient rule
With $u = x$, $u' = 1$, $v = 1+x^{2}$, $v' = 2x$: \[\frac{d}{dx}\!\left(\frac{x}{1+x^{2}}\right) = \frac{1\cdot\bigl(1+x^{2}\bigr) - x\cdot 2x}{\bigl(1+x^{2}\bigr)^{2}} = \frac{1+x^{2}-2x^{2}}{\bigl(1+x^{2}\bigr)^{2}} = \frac{1-x^{2}}{\bigl(1+x^{2}\bigr)^{2}}.\]
Step 3: Combine over a common denominator
Write the first derivative with denominator $\bigl(1+x^{2}\bigr)^{2}$ and subtract: \[y' = \frac{1+x^{2}}{\bigl(1+x^{2}\bigr)^{2}} - \frac{1-x^{2}}{\bigl(1+x^{2}\bigr)^{2}} = \frac{\bigl(1+x^{2}\bigr)-\bigl(1-x^{2}\bigr)}{\bigl(1+x^{2}\bigr)^{2}} = \frac{2x^{2}}{\bigl(1+x^{2}\bigr)^{2}}.\]
Step 4: Verification
\[x = 1:\quad y' = \frac{2}{4} = \frac12; \qquad \frac{1}{2} - \frac{1-1}{4} = \frac12 - 0 = \frac12 \;\;\checkmark\]
\[x = 0:\quad y'(0) = 0, \text{ and } y(0) = 0 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = \dfrac{2x^{2}}{\bigl(1+x^{2}\bigr)^{2}}\,}\]
Remark
Remark. Since $y' \geq 0$ everywhere, with equality only at $x = 0$, the function is increasing on all of $\mathbb{R}$; the origin is a stationary inflection, not an extremum. Read backwards, the computation says that $\dfrac{1}{2}\left(\arctan x - \dfrac{x}{1+x^{2}}\right)$ is an antiderivative of $\dfrac{x^{2}}{\bigl(1+x^{2}\bigr)^{2}}$.