Question

Differentiate the given function with respect to x using first principal

$f(x) = (x + 3)(x - 2)$
​

Answer 1

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

Given function is

$\rm \: f(x) = (x + 3)(x - 2)$

$\rm \: = {x}^{2} + 3x - 2x - 6$

$\rm \: = {x}^{2} + x - 6$

So,

$\rm \: f(x) = {x}^{2} + x - 6$

Now,

$\rm \: f(x + h) = {(x + h)}^{2} + x + h - 6$

By definition of First Principal, we have

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

$\rm \: = \: \displaystyle\lim_{h \to 0}\rm \frac{ {(x + h)}^{2} + x + h - 6 - ({x}^{2} + x - 6)}{h}$

$\rm \: = \: \displaystyle\lim_{h \to 0}\rm \frac{ {x}^{2} + {h}^{2} + 2xh + x + h - 6 - {x}^{2} - x + 6}{h}$

$\rm \: = \: \displaystyle\lim_{h \to 0}\rm \frac{{h}^{2} + 2xh + h}{h}$

$\rm \: = \: \displaystyle\lim_{h \to 0}\rm \frac{h(h + 2x + 1)}{h}$

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

$\rm \: = \: 2x + 1$

Hence,

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

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Additional Information

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