Find dy for y = (3x²-x)^(3/2)

Find $dy$ for $y = \bigl(3x^{2}-x\bigr)^{3/2}$.

Step 1: Differentiate by the chain rule

With outer function $u^{3/2}$ and inner function $u = 3x^{2}-x$: \[y' = \frac{3}{2}\bigl(3x^{2}-x\bigr)^{1/2}\cdot\frac{d}{dx}\bigl(3x^{2}-x\bigr) = \frac{3}{2}\bigl(3x^{2}-x\bigr)^{1/2}\bigl(6x-1\bigr).\]

Step 2: Form the differential

The differential is simply the derivative multiplied by $dx$: $dy = y'\,dx = \frac{3}{2}\bigl(6x-1\bigr)\sqrt{3x^{2}-x}\;dx$.

Step 3: Domain

The half-integer power requires $3x^{2}-x = x(3x-1) \ge 0$, i.e. $x\le 0$ or $x\ge\tfrac13$. At the two endpoints $x=0$ and $x=\tfrac13$ the square root vanishes, so $dy = 0$ there and the graph meets the $x$-axis with a horizontal tangent.

Step 4: Verification

\[x = 1:\quad 3x^{2}-x = 2,\ y = 2^{3/2} \approx 2.828427, \qquad y' = \frac{3}{2}\cdot 5\cdot\sqrt{2} \approx 10.606602\]

\[\frac{y(1.0001)-y(0.9999)}{0.0002} \approx 10.60661 \;\;\checkmark\]

\[\text{Alternative route: } y^{2} = (3x^{2}-x)^{3} \Rightarrow 2yy' = 3(3x^{2}-x)^{2}(6x-1) \Rightarrow y' = \tfrac{3}{2}(6x-1)\sqrt{3x^{2}-x} \;\;\checkmark\]