Using the definition of derivative, find $f^{\prime }(2)$ for $f(x)=\sqrt{5x-3}$.

Step-by-step solution

1. Write the limit definition: $f^{\prime }(2)={\lim }_{h\to 0}\frac{\sqrt{5(2+h)-3}-\sqrt{5\cdot 2-3}}{h}$.
2. Simplify the radicals: $\sqrt{5(2+h)-3}=\sqrt{10+5h-3}=\sqrt{7+5h}$ and $\sqrt{5\cdot 2-3}=\sqrt{7}$.
3. Rationalise the numerator by multiplying and dividing by $\sqrt{7+5h}+\sqrt{7}$.
4. The expression becomes ${\lim }_{h\to 0}\frac{(7+5h)-7}{h(\sqrt{7+5h}+\sqrt{7})}={\lim }_{h\to 0}\frac{5h}{h(\sqrt{7+5h}+\sqrt{7})}$.
5. Cancel $h$ and take the limit $h\to 0$: $\frac{5}{\sqrt{7}+\sqrt{7}}=\frac{5}{2\sqrt{7}}$.

Concept tested: Derivative by definition

Published 9 September 2026 · How this solution was prepared and checked