Find y' = lim as Δ x to 0 (Δ y)/(Δ x) for y = x³

Find $y' = \displaystyle\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}$ for $y = x^{3}$.

Step 1: Compute the incremented function value

\[ y + \Delta y = (x + \Delta x)^{3}. \]

Expanding with the binomial formula: $(x+\Delta x)^{3} = x^{3} + 3x^{2}\,\Delta x + 3x\,(\Delta x)^{2} + (\Delta x)^{3}$.

Step 2: Find the increment $\Delta y$

Subtract $y = x^{3}$: $\Delta y = 3x^{2}\,\Delta x + 3x\,(\Delta x)^{2} + (\Delta x)^{3}$.

Step 3: Form the difference quotient

Every term contains the factor $\Delta x$, so dividing by $\Delta x \neq 0$ gives

\[ \frac{\Delta y}{\Delta x} = 3x^{2} + 3x\,\Delta x + (\Delta x)^{2}. \]

Step 4: Take the limit

The last two terms vanish as $\Delta x \to 0$: \[y' = \lim_{\Delta x \to 0}\frac{\Delta y}{\Delta x} = \lim_{\Delta x \to 0}\bigl(3x^{2} + 3x\,\Delta x + (\Delta x)^{2}\bigr) = 3x^{2}.\]

Step 5: Verification

By the power rule $(x^{n})' = n x^{n-1}$ with $n = 3$: $y' = 3x^{3-1} = 3x^{2} \;\;\checkmark$.

\[x = 2,\ \Delta x = 0.001:\quad \frac{\Delta y}{\Delta x} = 3(2)^{2} + 3(2)(0.001) + (0.001)^{2} = 12.006001 \approx 12 = 3\cdot 2^{2} \;\;\checkmark\]