Find y' for y = nth root of x, where n in N and x > 0

Find $y'$ for $y = \sqrt[n]{x}$, where $n \in \mathbb{N}$ and $x \gt 0$.

Step 1: Rewrite as a power function

\[ y = \sqrt[n]{x} = x^{1/n}. \]

Step 2: Apply the power rule

Using $\bigl(x^{\alpha}\bigr)' = \alpha x^{\alpha-1}$ with $\alpha = \dfrac{1}{n}$: $y' = \frac{1}{n}\,x^{\,\frac{1}{n}-1}$.

Step 3: Simplify the exponent

\[ \frac{1}{n} - 1 = \frac{1-n}{n}, \qquad\text{so}\qquad y' = \frac{1}{n}\,x^{\,\frac{1-n}{n}}. \]

Step 4: Rewrite in radical form

Since $x^{(1-n)/n} = \dfrac{x^{1/n}}{x} = \dfrac{1}{x^{(n-1)/n}}$: \[y' = \frac{1}{n}\cdot\frac{\sqrt[n]{x}}{x} = \frac{1}{n\,x^{(n-1)/n}} = \frac{1}{n\,\sqrt[n]{x^{\,n-1}}}.\]

Step 5: Verification

Via the inverse function. Let $y = x^{1/n}$, so that $x = y^{n}$. Differentiating this relation with respect to $x$:

\[1 = n\,y^{n-1}\,\frac{dy}{dx} \;\Longrightarrow\; \frac{dy}{dx} = \frac{1}{n\,y^{n-1}} = \frac{1}{n\bigl(x^{1/n}\bigr)^{n-1}} = \frac{1}{n\,x^{(n-1)/n}} \;\;\checkmark\]

Special cases.

\[\begin{aligned} n = 2: &\quad \bigl(\sqrt{x}\bigr)' = \tfrac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}} \;\;\checkmark \\[4pt] n = 3: &\quad \bigl(\sqrt[3]{x}\bigr)' = \tfrac{1}{3}x^{-2/3} = \frac{1}{3\sqrt[3]{x^{2}}} \;\;\checkmark \end{aligned}\]

The case $n = 2$ evaluated at $x = 1$ gives $\tfrac12$, agreeing with Problem 6.