Find y' for y = x^n, where n in N

Find $y'$ for $y = x^{n}$, where $n \in \mathbb{N}$.

Step 1: Write the difference quotient

\[ \frac{\Delta y}{\Delta x} = \frac{(x+\Delta x)^{n} - x^{n}}{\Delta x}. \]

Step 2: Expand by the binomial theorem

\[ (x+\Delta x)^{n} = x^{n} + nx^{n-1}\Delta x + \binom{n}{2}x^{n-2}(\Delta x)^{2} + \cdots + (\Delta x)^{n}. \]

Subtracting $x^{n}$: $\Delta y = nx^{n-1}\Delta x + \binom{n}{2}x^{n-2}(\Delta x)^{2} + \cdots + (\Delta x)^{n}$.

Step 3: Form the difference quotient

Every term contains at least one factor $\Delta x$, so dividing by $\Delta x \neq 0$ gives

\[ \frac{\Delta y}{\Delta x} = nx^{n-1} + \binom{n}{2}x^{n-2}\Delta x + \cdots + (\Delta x)^{n-1}. \]

Every term except the first still contains a factor $\Delta x$.

Step 4: Take the limit

As $\Delta x \to 0$ all terms carrying a factor $\Delta x$ vanish, and only the first survives: $y' = \lim_{\Delta x\to 0}\frac{\Delta y}{\Delta x} = nx^{n-1}$.

Step 5: Verification — special cases

\[\begin{aligned} n = 1: &\quad (x)' = 1\cdot x^{0} = 1 \;\;\checkmark \\[4pt] n = 2: &\quad (x^{2})' = 2x \;\;\checkmark \\[4pt] n = 3: &\quad (x^{3})' = 3x^{2} \;\;\checkmark \end{aligned}\]

The case $n = 3$ agrees with the first-principles computation of Problem 10.