Question

find the derivative of the following function using the definition.
$\frac{1}{3x + 2}$
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm \: f(x) = \dfrac{1}{3x + 2}$

So,

$\rm \: f(x + h) = \dfrac{1}{3(x + h) + 2} = \dfrac{1}{3x + 3h + 2}$

Now, By definition of differentiation, we have

$\rm \: f'(x) = \displaystyle\lim_{h \to 0}\rm \dfrac{f(x + h) - f(x)}{h}$

So, on substituting the values, we get

$\rm \: = \: \displaystyle\lim_{h \to 0}\rm \bigg(\dfrac{1}{3x + 3h + 2} - \dfrac{1}{3x + 2} \bigg) \dfrac{1}{h}$

$\rm \: = \: \displaystyle\lim_{h \to 0}\rm \bigg(\dfrac{3x + 2 - (3x + 3h + 2)}{(3x + 3h + 2)(3x + 2)} \bigg) \dfrac{1}{h}$

$\rm \: = \: \displaystyle\lim_{h \to 0}\rm \bigg(\dfrac{3x + 2 - 3x - 3h - 2}{(3x + 3h + 2)(3x + 2)} \bigg) \dfrac{1}{h}$

$\rm \: = \: \displaystyle\lim_{h \to 0}\rm \bigg(\dfrac{- 3h}{(3x + 3h + 2)(3x + 2)} \bigg) \dfrac{1}{h}$

$\rm \: = \: \displaystyle\lim_{h \to 0}\rm \bigg(\dfrac{- 3}{(3x + 3h + 2)(3x + 2)} \bigg)$

$\rm \: = \: \dfrac{ - 3}{(3x + 2)(3x + 2)}$

$\rm \: = \: \dfrac{ - 3}{ {(3x + 2)}^{2} }$

Hence,

$\rm\implies \: \boxed{ \rm{ \:\: \dfrac{d}{dx}\bigg( \dfrac{1}{3x + 2}\bigg) \: = \: \dfrac{ - 3}{ {(3x + 2)}^{2} } \: \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$