Find y' for y = (x²-2) sqrt(x)

Find $y'$ for $y = \bigl(x^{2}-2\bigr)\sqrt{x}$.

Step 1: Method 1: expand first

Multiplying out and collecting powers of $x$: $y = x^{2}\cdot x^{1/2} - 2x^{1/2} = x^{5/2} - 2x^{1/2}$.

Differentiate term by term with the power rule: $y' = \frac{5}{2}x^{3/2} - 2\cdot\frac{1}{2}x^{-1/2} = \frac{5}{2}x^{3/2} - x^{-1/2}$.

Putting both terms over the common denominator $2\sqrt{x}$: $y' = \frac{5x^{2} - 2}{2\sqrt{x}}$.

Step 2: Method 2: the product rule

For $y = uv$ recall $y' = u'v + uv'$. Here

\[ u = x^{2}-2,\quad u' = 2x, \qquad v = \sqrt{x},\quad v' = \frac{1}{2\sqrt{x}}, \]

so that

\[ y' = 2x\sqrt{x} + \bigl(x^{2}-2\bigr)\cdot\frac{1}{2\sqrt{x}}. \]

Step 3: Bring to a common denominator

Since $2x\sqrt{x} = \dfrac{4x^{2}}{2\sqrt{x}}$: $y' = \frac{4x^{2} + x^{2} - 2}{2\sqrt{x}} = \frac{5x^{2}-2}{2\sqrt{x}}$.

Both methods give the same answer, as they must.

Step 4: Verification

\[x = 1:\quad y' = \frac{5-2}{2} = \frac{3}{2}; \qquad \frac{y(1.001)-y(0.999)}{0.002} \approx 1.5 \;\;\checkmark\]

\[y' = 0 \;\Longleftrightarrow\; x^{2} = \frac{2}{5} \;\Longleftrightarrow\; x = \sqrt{0.4} \approx 0.632, \text{ the minimum of } y \;\;\checkmark\]