Find y' for y = (x²)/(x²+1)
Find $y'$ for $y = \dfrac{x^{2}}{x^{2}+1}$.
Explanation of Method
Apply the quotient rule and watch the leading terms cancel.
Step 1: Identify the parts and apply the quotient rule
For $y = \dfrac{u}{v}$ recall $y' = \dfrac{u'v - uv'}{v^{2}}$. Here
\[ u = x^{2},\quad u' = 2x, \qquad v = x^{2}+1,\quad v' = 2x. \]
Step 2: Substitute and simplify
\[ y' = \frac{2x\bigl(x^{2}+1\bigr) - x^{2}\cdot 2x}{\bigl(x^{2}+1\bigr)^{2}} = \frac{2x^{3}+2x-2x^{3}}{\bigl(x^{2}+1\bigr)^{2}} = \frac{2x}{\bigl(x^{2}+1\bigr)^{2}}. \]
The cubic terms cancel, leaving a strikingly simple numerator.
Step 3: Verification
An independent route: divide out first, writing $y = 1 - \dfrac{1}{x^{2}+1}$. Then
\[y' = 0 - \left(-\frac{2x}{\bigl(x^{2}+1\bigr)^{2}}\right) = \frac{2x}{\bigl(x^{2}+1\bigr)^{2}} \;\;\checkmark\]
\[x = 1:\quad y'(1) = \frac{2}{4} = \frac{1}{2}; \qquad \frac{y(1.001)-y(0.999)}{0.002} \approx 0.5 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = \dfrac{2x}{\bigl(x^{2}+1\bigr)^{2}}\,}\]
Remark
Remark. The sign of $y'$ is the sign of $x$, so the function decreases on $(-\infty,0)$, has a minimum $y(0) = 0$, and increases on $(0,\infty)$, approaching the horizontal asymptote $y = 1$. The denominator never vanishes, so the formula holds for every real $x$.