Q.Find the derivative of $f(x)=\sqrt{x}$ ($x>0$) from first principles.

By definition, the derivative from first principles is $f'(x) = \displaystyle\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}$. Factor out the common term and use the standard limit $\lim_{t\to0}\dfrac{a^t-1}{t} = \ln a$.

Given: $y = f(x) = a^{2x+5}$

Step 1 — set up the definition:

$f'(x) = \lim_{h\to0} \frac{a^{2(x+h)+5} - a^{2x+5}}{h}$

Step 2 — factor out $a^{2x+5}$:

$= \lim_{h\to0} \frac{a^{2x+5}\left(a^{2h} - 1\right)}{h} = a^{2x+5}\lim_{h\to0}\frac{a^{2h}-1}{h}$

Step 3 — evaluate the limit. Let $t = 2h$; as $h \to 0$, $t\to 0$ and $h = t/2$: …