Q.Differentiate $f(x) = 3x^{2}$ 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 $h = t/2$:

$\lim_{h\to0}\frac{a^{2h}-1}{h} = \lim_{t\to0}\frac{a^t-1}{t/2} = 2\lim_{t\to0}\frac{a^t-1}{t} = 2\ln a$

(using the standard result $\lim_{t\to0}\dfrac{a^t-1}{t} = \ln a$).

Step 4 — combine:

$f'(x) = a^{2x+5}\cdot 2\ln a = 2\ln a \cdot a^{2x+5}$