Find f'(1) by definition for f(x) = sqrt(x)

Find $f'(1)$ by definition for $f(x) = \sqrt{x}$.

Step 1: Write the definition of the derivative at $x = 1$

\[ f'(1) = \lim_{\Delta x \to 0} \frac{f(1 + \Delta x) - f(1)}{\Delta x}. \]

Step 2: Compute the increment $\Delta y$

\[ f(1 + \Delta x) = \sqrt{1 + \Delta x}, \qquad f(1) = \sqrt{1} = 1, \]

so that

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

Step 3: Rationalise the difference quotient

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

Cancelling $\Delta x \neq 0$: $\frac{\Delta y}{\Delta x} = \frac{1}{\sqrt{1+\Delta x}+1}$.

Step 4: Take the limit

\[ f'(1) = \lim_{\Delta x \to 0} \frac{1}{\sqrt{1+\Delta x}+1} = \frac{1}{\sqrt{1+0}+1} = \frac{1}{1+1} = \frac{1}{2}. \]

Step 5: Verification

Writing $f(x) = x^{1/2}$ and using the power rule $(x^{a})' = a x^{a-1}$: $f'(x) = \tfrac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$.

\[f'(1) = \frac{1}{2\sqrt{1}} = \frac{1}{2} \;\;\checkmark\]