Find the derivation of each of the following from the first principle:

Solution:

Let $f(x)=3 x^{2}+2 x-5$

We need to find the derivative of f(x) i.e. f’(x)

We know that,

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$ …(i)

$f(x)=3 x^{2}+2 x-5$

$f(x+h)=3(x+h)^{2}+2(x+h)-5$

$=3\left(x^{2}+h^{2}+2 x h\right)+2 x+2 h-5$

$\left[\because(a+b)^{2}=a^{2}+b^{2}+2 a b\right]$

$=3 x^{2}+3 h^{2}+6 x h+2 x+2 h-5$

Putting values in (i), we get

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{3 x^{2}+3 h^{2}+6 x h+2 x+2 h-5-\left(3 x^{2}+2 x-5\right)}{h}$

$=\lim _{h \rightarrow 0} \frac{3 x^{2}+3 h^{2}+6 x h+2 x+2 h-5-3 x^{2}-2 x+5}{h}$

$=\lim _{h \rightarrow 0} \frac{3 h^{2}+6 x h+2 h}{h}$

$=\lim _{h \rightarrow 0} 3 h+6 x+2$

Putting h = 0, we get

$f^{\prime}(x)=3(0)+6 x+2$

$=6 x+2$

Hence, $f^{\prime}(x)=6 x+2$