Differentiate f(x) = x/(x^2 + 1).

Step-by-step solution

1. 1

   Set up the quotient rule. With $u=x$ and $v=x^2+1$,

   $f'=\frac{u'v-uv'}{v^2},\qquad u'=1,\qquad v'=2x$

   The order in the numerator matters: it is derivative of top times bottom minus top times derivative of bottom, never the reverse. Note also that $v=x^2+1\ge1>0$ everywhere, so $f$ is defined and differentiable on all of $\mathbb{R}$.

2. 2

   Substitute into the formula.

   $f'(x)=\frac{(1)\left(x^2+1\right)-(x)(2x)}{\left(x^2+1\right)^2}$

3. 3

   Simplify the numerator.

   $x^2+1-2x^2=1-x^2$

   so

   $f'(x)=\frac{1-x^2}{\left(x^2+1\right)^2}$

   The $x^2$ terms partially cancel — which is what turns a messy expression into a clean one.

4. 4

   Read the turning points straight off the answer. The denominator is always positive, so the sign of $f'$ is the sign of $1-x^2=(1-x)(1+x)$. That is positive on $(-1,1)$ and negative outside, giving a local minimum at $x=-1$ (where $f=-\tfrac12$) and a local maximum at $x=1$ (where $f=\tfrac12$).

5. 5

   Note the end behaviour. As $|x|\to\infty$ the denominator grows like $x^2$ while the numerator grows like $x$, so $f(x)\to0$: the $x$-axis is a horizontal asymptote in both directions, and $\pm\tfrac12$ are the global extreme values.

6. 6

   Verify numerically. Comparing the closed form against symmetric difference quotients of step $10^{-7}$: at $x=0.4$ both give $0.624257$; at $x=-1.7$ both give $-0.124900$; at $x=3.3$ both give $-0.069957$ ✓.