Differentiate f(x) = x/(x^2 + 1) using the quotient rule.

Step-by-step solution

1. 1

   Recognise the shape and pick the rule. The function is a single fraction whose numerator and denominator both contain $x$, so neither the constant-multiple rule nor the power rule applies directly. The quotient rule is built for exactly this:

   $\left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^{2}}$

2. 2

   Name the two parts and differentiate each.

   $u=x\ \Longrightarrow\ u'=1,\qquad v=x^{2}+1\ \Longrightarrow\ v'=2x$

   Writing them out before substituting is what stops the classic mistake of swapping the order in the numerator — the rule is $u'v$ minus $uv'$, and it is not symmetric.

3. 3

   Substitute into the rule.

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

4. 4

   Simplify the numerator only. The denominator is already in its most useful factored form, so leave it alone:

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

   giving

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

5. 5

   Read the answer for a free sanity check. The numerator $1-x^{2}$ is positive on $(-1,1)$ and negative outside it, so $f$ rises between $x=-1$ and $x=1$ and falls elsewhere — matching the fact that $f$ has its maximum $\tfrac12$ at $x=1$ and its minimum $-\tfrac12$ at $x=-1$.

6. 6

   Verify numerically. Central differences with $h=10^{-6}$ give $f'(0.3)\approx 0.7659288$ against the formula's $0.7659288$, and $f'(2)\approx -0.1200000$ against $-3/25=-0.12$.