Find y' for y = a/(sqrt(x²)) - b/(x cube root of x)

Find $y'$ for $y = \dfrac{a}{\sqrt{x^{2}}} - \dfrac{b}{x\sqrt[3]{x}}$, where $a$ and $b$ are constants and $x \gt 0$.

Step 1: Rewrite each term as a power of $x$

First term. In general $\sqrt{x^{2}} = |x|$; for $x \gt 0$ this is simply $x$, so

\[ \frac{a}{\sqrt{x^{2}}} = \frac{a}{x} = a\,x^{-1}. \]

Second term. Collecting the powers in the denominator,

\[ x\,\sqrt[3]{x} = x\cdot x^{1/3} = x^{1+\frac13} = x^{4/3}, \qquad\text{so}\qquad \frac{b}{x\sqrt[3]{x}} = b\,x^{-4/3}. \]

The function becomes a sum of two power terms: $y = a\,x^{-1} - b\,x^{-4/3}$.

Step 2: Differentiate term by term

Apply $\bigl(x^{\alpha}\bigr)' = \alpha x^{\alpha-1}$ to each term: \[y' = a\cdot(-1)\,x^{-2} - b\cdot\left(-\frac{4}{3}\right)x^{-\frac43-1} = -a\,x^{-2} + \frac{4b}{3}\,x^{-7/3}.\]

Step 3: Rewrite in explicit form

Since $x^{7/3} = x^{2}\cdot x^{1/3} = x^{2}\sqrt[3]{x}$: \[y' = -\frac{a}{x^{2}} + \frac{4b}{3\,x^{7/3}} = -\frac{a}{x^{2}} + \frac{4b}{3\,x^{2}\sqrt[3]{x}}.\]

Step 4: Verification

\[b = 0:\quad y = \frac{a}{x} \;\Longrightarrow\; y' = -\frac{a}{x^{2}}, \text{ agreeing with Problem 13} \;\;\checkmark\]

\[a=0,\ b=1,\ x=1:\quad y' = \frac{4}{3} \approx 1.3333; \qquad \frac{y(1.001)-y(1)}{0.001} \approx 1.3318 \;\;\checkmark\]