Find the approximate increment of y = sqrt(x) when x

Find the approximate increment of $y = \sqrt{x}$ when $x$ changes from $100$ to $101$.

Step 1: The approximation being used

For a small change $\Delta x$ the true increment $\Delta y = f(x_{0}+\Delta x)-f(x_{0})$ is well approximated by the differential: $\Delta y \approx dy = f'(x_{0})\,\Delta x$.

Geometrically, the curve is replaced by its tangent line at $x_{0}$.

Step 2: Apply it

Here $x_{0} = 100$, $\Delta x = 1$, and $y' = \dfrac{1}{2\sqrt{x}}$: \[dy = \frac{1}{2\sqrt{100}}\cdot 1 = \frac{1}{20} = 0.05.\]

Step 3: How good is it?

The exact increment is

\[ \Delta y = \sqrt{101}-\sqrt{100} = 10.0498756 - 10 = 0.0498756. \]

\[ \bigl|\Delta y - dy\bigr| = 0.0001244, \qquad \frac{\bigl|\Delta y-dy\bigr|}{\Delta y} \approx 0.25\%. \]

So the tangent-line estimate overshoots by about one part in four hundred — the square root is concave, so its tangent lies above the curve and $dy > \Delta y$ always.

Step 4: Verification

\[10.05^{2} = 101.0025 \quad\text{(slightly more than 101, confirming the overshoot)} \;\;\checkmark\]

\[\text{Second-order term: } -\frac{1}{8x_{0}^{3/2}}(\Delta x)^{2} = -\frac{1}{8000} = -0.000125 \approx \Delta y - dy \;\;\checkmark\]

\[\text{Refined estimate: } 10+0.05-0.000125 = 10.049875 \text{ versus } \sqrt{101} = 10.0498756 \;\;\checkmark\]