Q.Find the derivative of $f(x) = \sqrt{x}$ from first principles.

The difference quotient here involves a difference of square roots, which does not simplify by ordinary factoring — instead we rationalise by multiplying by the conjugate before letting $h\to0$.

Step 1 — Write the difference quotient. $f'(x) = \displaystyle\lim_{h\to0}\dfrac{\sqrt{x+h}-\sqrt x}{h}$, with $f(x)=\sqrt x$.

Step 2 — Rationalise. The numerator is a difference of square roots, which does not cancel with $h$ directly — multiply numerator and denominator by the conjugate $\sqrt{x+h}+\sqrt x$:

Step 3 — Cancel $h$. $= \dfrac{1}{\sqrt{x+h}+\sqrt x}$ (valid since $h\neq0$ during the limit process).

Step 4 — Let $h\to0$. As $h\to0$, $\sqrt{x+h}\to\sqrt x$, so …

Trying to cancel the $h$ directly from $\sqrt{x+h}-\sqrt x$ without rationalising first is the central error here — square roots do not distribute or cancel that way, and rationalisa …