Question

$\large {\dag \; {\underline{\underline{\red{\pmb{\frak{ \; Question \; :- }}}}}}}$
⇢Find the derivatives of the Following functions which respect to x by using First principle:-

$\bf{1). \: \: \: {x}^{ \frac{3}{2} } }$
$\bf{b). \: \: \frac{1}{ax + b} }$

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Answer 1

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

Given function is

$\rm \: f(x) = {\bigg(x\bigg) }^{\dfrac{3}{2} }$

So,

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

By using Definition of First Principal,

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

So, on substituting the values, we get

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

can be further rewritten as

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

We know,

$\boxed{\sf{ \: \: \displaystyle\lim_{x \to a}\rm \frac{ {x}^{n} - {a}^{n} }{x - a} \: = \: {na}^{n - 1} \: \: }}$

So, using this result, we get

$\rm \: = \: \dfrac{3}{2} {\bigg(x\bigg) }^{\dfrac{3}{2} - 1}$

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

$\rm \: = \: \dfrac{3}{2} {\bigg(x\bigg) }^{\dfrac{1}{2}}$

$\rm \: = \: \dfrac{3}{2} \sqrt{x}$

So,

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

$\large\underline{\sf{Solution-(ii)}} \:$

Given function is

$\rm \: f(x) = \dfrac{1}{ax + b}$

So,

$\rm \: f(x) = \dfrac{1}{a(x + h)+ b} = \dfrac{1}{ax + ah + b}$

By, using definition of First Principal, we have

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

So, on substituting the values, we get

$\rm \: = \: \displaystyle\lim_{h \to 0}\rm \frac{1}{h}\bigg(\dfrac{1}{ax + ah+ b} - \dfrac{1}{ax + b}\bigg)$

$\rm \: = \: \displaystyle\lim_{h \to 0}\rm \frac{1}{h}\bigg(\dfrac{ax + b - ax - ah - b}{(ax + ah+ b)(ax + b)} \bigg)$

$\rm \: = \: \displaystyle\lim_{h \to 0}\rm \frac{1}{h}\bigg(\dfrac{ - ah}{(ax + ah+ b)(ax + b)} \bigg)$

$\rm \: = \: \displaystyle\lim_{h \to 0}\rm \bigg(\dfrac{ - a}{(ax + ah+ b)(ax + b)} \bigg)$

$\rm \: = \: \dfrac{ - a}{(ax + b)(ax + b)}$

$\rm \: = \: - \: \dfrac{a}{(ax + b)^{2} }$

Hence,

$\rm\implies \:\boxed{\sf{ \: \: \rm \:\dfrac{d}{dx} \: \dfrac{1}{ax + b} \: = \: - \: \dfrac{a}{(ax + b)^{2} } \: \: }}$

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Additional Information

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