Question

Differentiate the following function with respect to x using first principal

$f(x) = \frac{2x + 1}{3x + 4}$
​

Answer 1

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

Given function is

$\rm \: f(x) = \dfrac{2x + 1}{3x + 4}$

So,

$\rm \: f(x + h) = \dfrac{2(x + h) + 1}{3(x + h) + 4} = \dfrac{2x + 2h + 1}{3x + 3h + 4}$

By using Definition of First Principal, we have

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

On substituting the values, we get

$\rm \: f'(x) = \displaystyle\lim_{h \to 0}\rm \frac{1}{h}\bigg(\dfrac{2x + 2h + 1}{3x + 3h + 4} - \dfrac{2x + 1}{3x + 4} \bigg)$

$\rm \: = \displaystyle\lim_{h \to 0}\rm \frac{1}{h}\bigg(\dfrac{(2x + 2h + 1)(3x + 4) - (2x + 1)(3x + 3h + 4)}{(3x + 3h + 4)(3x + 4)} \bigg)$

$\rm \: = \displaystyle\lim_{h \to 0}\rm \bigg(\dfrac{ {6x}^{2} + 6xh + 3x + 8x + 8h + 4 - {6x}^{2} - 6hx - 8x - 3x - 3h - 4 }{h(3x + 3h + 4)(3x + 4)} \bigg)$

$\rm \: = \displaystyle\lim_{h \to 0}\rm \bigg(\dfrac{ 5h }{h(3x + 3h + 4)(3x + 4)} \bigg)$

$\rm \: = \displaystyle\lim_{h \to 0}\rm \dfrac{ 5 }{(3x + 3h + 4)(3x + 4)}$

$\rm \: = \dfrac{ 5 }{(3x + 4)(3x + 4)}$

$\rm \: = \dfrac{ 5 }{(3x + 4)^{2} }$

Hence,

$\rm\implies \:\boxed{\sf{ \:\rm \: \dfrac{d}{dx}\bigg(\dfrac{2x + 1}{3x + 4}\bigg) = \frac{5}{ {(3x + 4)}^{2} } \: \: }}$

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Short Cut Trick

$\: \: \boxed{\sf{ \:\rm \: \: \: \frac{d}{dx} \bigg(\frac{ax + b}{cx + d}\bigg) = \frac{(ad - bc)}{(cx + d)^{2} } \: \: \: \: }}$

Additional Information

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