Find y' for y = (x²-1) (x²-4) (x²-9)
Find $y'$ for $y = \bigl(x^{2}-1\bigr)\bigl(x^{2}-4\bigr)\bigl(x^{2}-9\bigr)$.
Explanation of Method
Extend the product rule to three factors.
Step 1: State the product rule for three factors
Applying the two-factor rule twice gives, for $y = uvw$, $y' = u'vw + uv'w + uvw'$.
Set $u = x^{2}-1$, $v = x^{2}-4$, $w = x^{2}-9$; each factor has the same derivative $u' = v' = w' = 2x$.
Step 2: Substitute
\[ y' = 2x\bigl(x^{2}-4\bigr)\bigl(x^{2}-9\bigr) + \bigl(x^{2}-1\bigr)2x\bigl(x^{2}-9\bigr) + \bigl(x^{2}-1\bigr)\bigl(x^{2}-4\bigr)2x. \]
Step 3: Factor out $2x$
\[ y' = 2x\Bigl[\bigl(x^{2}-4\bigr)\bigl(x^{2}-9\bigr) + \bigl(x^{2}-1\bigr)\bigl(x^{2}-9\bigr) + \bigl(x^{2}-1\bigr)\bigl(x^{2}-4\bigr)\Bigr]. \]
Step 4: Expand and add the three brackets
\[ \begin{aligned} \bigl(x^{2}-4\bigr)\bigl(x^{2}-9\bigr) &= x^{4} - 13x^{2} + 36, \\ \bigl(x^{2}-1\bigr)\bigl(x^{2}-9\bigr) &= x^{4} - 10x^{2} + 9, \\ \bigl(x^{2}-1\bigr)\bigl(x^{2}-4\bigr) &= x^{4} - 5x^{2} + 4, \\[4pt] \text{sum} &= 3x^{4} - 28x^{2} + 49. \end{aligned} \]
Step 5: Verification
Expanding $y$ completely gives $y = x^{6} - 14x^{4} + 49x^{2} - 36$; differentiating that polynomial directly: $y' = 6x^{5} - 56x^{3} + 98x = 2x\bigl(3x^{4} - 28x^{2} + 49\bigr) \;\;\checkmark$.
\[x = 1:\quad y'(1) = 2(3-28+49) = 48; \qquad \frac{y(1.001)-y(0.999)}{0.002} \approx 48.0 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = 2x\bigl(3x^{4} - 28x^{2} + 49\bigr)\,}\]
Remark
Remark. The function is even, so its derivative is odd — visible in the factor $2x$. Since $y$ vanishes at the six points $x = \pm 1, \pm 2, \pm 3$, Rolle's theorem guarantees a zero of $y'$ between consecutive ones; indeed $3x^{4}-28x^{2}+49 = 0$ gives $x^{2} = \dfrac{14\pm\sqrt{49}}{3}$, i.e. $x^{2} = 7$ or $x^{2} = \tfrac73$, together with the root $x = 0$.