Find y' = lim as Δ x to 0 (Δ y)/(Δ x) for y = sqrt(1+2x)

Find $y' = \displaystyle\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}$ for $y = \sqrt{1+2x}$.

Step 1: Compute the incremented function value

\[ y + \Delta y = \sqrt{1 + 2(x + \Delta x)} = \sqrt{1 + 2x + 2\,\Delta x}. \]

Step 2: Find the increment $\Delta y$

\[ \Delta y = \sqrt{1+2x+2\,\Delta x} - \sqrt{1+2x}. \]

Step 3: Rationalise the difference quotient

The quotient

\[ \frac{\Delta y}{\Delta x} = \frac{\sqrt{1+2x+2\,\Delta x}-\sqrt{1+2x}}{\Delta x} \]

has the indeterminate form $\tfrac{0}{0}$. Multiply numerator and denominator by the conjugate $\sqrt{1+2x+2\,\Delta x}+\sqrt{1+2x}$: \[\frac{\Delta y}{\Delta x} = \frac{\bigl(\sqrt{1+2x+2\,\Delta x}\bigr)^{2} - \bigl(\sqrt{1+2x}\bigr)^{2}}{\Delta x\,\bigl(\sqrt{1+2x+2\,\Delta x}+\sqrt{1+2x}\bigr)}.\]

The numerator simplifies to

\[ (1+2x+2\,\Delta x)-(1+2x) = 2\,\Delta x, \]

so that, after cancelling $\Delta x \neq 0$, \[\frac{\Delta y}{\Delta x} = \frac{2\,\Delta x}{\Delta x\,\bigl(\sqrt{1+2x+2\,\Delta x}+\sqrt{1+2x}\bigr)} = \frac{2}{\sqrt{1+2x+2\,\Delta x}+\sqrt{1+2x}}.\]

Step 4: Take the limit

\[ y' = \lim_{\Delta x \to 0} \frac{2}{\sqrt{1+2x+2\,\Delta x}+\sqrt{1+2x}} = \frac{2}{\sqrt{1+2x}+\sqrt{1+2x}} = \frac{2}{2\sqrt{1+2x}} = \frac{1}{\sqrt{1+2x}}. \]

Step 5: Verification

By the chain rule with $u = 1+2x$: \[y' = \frac{1}{2\sqrt{u}}\cdot u' = \frac{1}{2\sqrt{1+2x}}\cdot 2 = \frac{1}{\sqrt{1+2x}} \;\;\checkmark\]

\[x = 0,\ \Delta x = 10^{-4}:\quad \frac{\sqrt{1.0002}-1}{10^{-4}} \approx 0.99995 \approx 1 = \frac{1}{\sqrt{1}} \;\;\checkmark\]