Using the definition of derivative, find $f^{\prime }(3)$ for $f(x)=\sqrt{2x+1}$.

Step-by-step solution

1. Write the definition: $f^{\prime }(3)={\lim }_{h\to 0}\frac{\sqrt{2(3+h)+1}-\sqrt{2\cdot 3+1}}{h}$.
2. Simplify the radicands: $\sqrt{2(3+h)+1}=\sqrt{7+2h}$ and $\sqrt{2\cdot 3+1}=\sqrt{7}$.
3. Rationalise the numerator: multiply numerator and denominator by $\sqrt{7+2h}+\sqrt{7}$.
4. The expression becomes ${\lim }_{h\to 0}\frac{(7+2h)-7}{h(\sqrt{7+2h}+\sqrt{7})}={\lim }_{h\to 0}\frac{2h}{h(\sqrt{7+2h}+\sqrt{7})}$.
5. Cancel $h$: $={\lim }_{h\to 0}\frac{2}{\sqrt{7+2h}+\sqrt{7}}$.
6. Let $h\to 0$: denominator $\to 2\sqrt{7}$, so the limit is $\frac{2}{2\sqrt{7}}=\frac{1}{\sqrt{7}}$.
7. Thus $f^{\prime }(3)=\frac{1}{\sqrt{7}}$.

Concept tested: Derivative from first principles

Published 7 September 2026 · How this solution was prepared and checked