Question

Find the derivative by first principle
(1) $3x^2+4$
(2) $\frac{1}{2x+3}$

Answer 1

(i) $f^1\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$

$=\underset{h\to 0}{lim}\frac{[{3(x+h)}^2+4]-[{3x}^2+4]}{h}$

$=\underset{h\to 0}{lim}\frac{\left[3(h^2+2xh)\right]}{h}$

$=\underset{h\to 0}{lim}3(h+2x)$

$=6x$

(ii) $f\left(x\right)=\frac{1}{2x+3}$

$f^1\left(x\right)=\underset{h\to 0}{lim}\frac{\frac{1}{2(x+h)+3}-\frac{1}{2x+3}}{h}=\underset{h\to 0}{lim}\frac{(2x+3)-[(2x+2h)+3]}{h(2x+3)(2x+2h+3)}$

$=\underset{h\to 0}{lim}\frac{(2x+3)-(2x+1h+3)}{(2x+3)h(2x+2h+3)}=\underset{h\to 0}{lim}\frac{-2h}{h(2x+3)(2x+2h+3)}$

$=\frac{-2}{{(2x+3)}^2}$