Find y' for y = x/(1-x²)
Find $y'$ for $y = \dfrac{x}{1-x^{2}}$.
Explanation of Method
A first application of the quotient rule.
Step 1: Identify the parts and apply the quotient rule
For $y = \dfrac{u}{v}$ the quotient rule reads $y' = \dfrac{u'v - uv'}{v^{2}}$. Here
\[ u = x,\quad u' = 1, \qquad v = 1-x^{2},\quad v' = -2x. \]
Step 2: Substitute and simplify
\[ y' = \frac{1\cdot\bigl(1-x^{2}\bigr) - x\cdot(-2x)}{\bigl(1-x^{2}\bigr)^{2}} = \frac{1-x^{2}+2x^{2}}{\bigl(1-x^{2}\bigr)^{2}} = \frac{1+x^{2}}{\bigl(1-x^{2}\bigr)^{2}}. \]
Note how the two $x^{2}$ terms combine rather than cancel, because of the minus sign in $v'$.
Step 3: Verification
\[x = 0:\quad y'(0) = \frac{1}{1} = 1, \text{ matching the tangent } y = x \text{ at the origin} \;\;\checkmark\]
\[x = 2:\quad y'(2) = \frac{5}{9} \approx 0.5556; \qquad \frac{y(2.001)-y(1.999)}{0.002} \approx 0.5556 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = \dfrac{1+x^{2}}{\bigl(1-x^{2}\bigr)^{2}}, \qquad x \neq \pm 1\,}\]
Remark
Remark. Numerator and denominator are both strictly positive wherever they are defined, so $y' \gt 0$ everywhere: the function increases on each of the three intervals $(-\infty,-1)$, $(-1,1)$ and $(1,\infty)$ separately, with vertical asymptotes at $x = \pm 1$.