Find y' for y = 1/((1-x²)^5) = (1-x²)^-5
Find $y'$ for $y = \dfrac{1}{\bigl(1-x^{2}\bigr)^{5}} = \bigl(1-x^{2}\bigr)^{-5}$.
Explanation of Method
A negative power handled by the chain rule rather than the quotient rule.
Step 1: Apply the chain rule
Writing the reciprocal as a negative power avoids the quotient rule entirely. With
\[ u = 1-x^{2}, \qquad u' = -2x, \qquad n = -5: \]
\[ y' = -5\bigl(1-x^{2}\bigr)^{-6}\cdot(-2x). \]
Step 2: Simplify
The two minus signs cancel: $y' = \frac{10x}{\bigl(1-x^{2}\bigr)^{6}}$.
Step 3: Verification
\[x = 0:\quad y'(0) = 0, \text{ and } y(0) = 1 \text{ is a local minimum} \;\;\checkmark\]
\[x = \tfrac12:\quad y' = \frac{5}{(3/4)^{6}} = \frac{5\cdot 4096}{729} \approx 28.09\]
\[\frac{y(0.5+10^{-5})-y(0.5-10^{-5})}{2\times 10^{-5}} \approx 28.09 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = \dfrac{10x}{\bigl(1-x^{2}\bigr)^{6}}, \qquad x \neq \pm 1\,}\]
Remark
Remark. Because the exponent $6$ is even, the denominator is positive wherever it is defined, so the sign of $y'$ is simply the sign of $x$. The function is even, its derivative odd, and both blow up at the vertical asymptotes $x = \pm 1$.