Math, asked by ItxAttitude, 12 hours ago

derivate
1 \div \sqrt{x} 1÷
x

by first principle

Answers

Answered by EmperorSoul

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

Given function is

$\rm :\longmapsto\:f(x) = \dfrac{1}{ \sqrt{x} }$

So,

$\rm :\longmapsto\:f(x + h) = \dfrac{1}{ \sqrt{x + h} }$

So, by definition of First Principle, we have

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

So, on substituting the values, we get

$\rm :\longmapsto\:f'(x) = \displaystyle\lim_{h \to 0}\rm \frac{1}{h}\bigg[\dfrac{1}{ \sqrt{x + h} } - \frac{1}{ \sqrt{x} } \bigg]$

$\rm :\longmapsto\:f'(x) = \displaystyle\lim_{h \to 0}\rm \frac{1}{h}\bigg[\dfrac{ \sqrt{x} - \sqrt{x + h} }{ \sqrt{x + h} \: \sqrt{x} }\bigg]$

$\rm \: = \: \displaystyle\lim_{h \to 0}\rm \frac{1}{ \sqrt{x + h} \: \sqrt{x} } \times \displaystyle\lim_{h \to 0}\rm \frac{ \sqrt{x} - \sqrt{x + h} }{h}$

$\rm \: = \: \dfrac{1}{ \sqrt{x} \: \sqrt{x} } \times \displaystyle\lim_{h \to 0}\rm \frac{ \sqrt{x} - \sqrt{x + h} }{h} \times \frac{ \sqrt{x} + \sqrt{x + h} }{ \sqrt{x} + \sqrt{x + h} }$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ (x + y)(x - y) = {x}^{2} - {y}^{2} \: }}} \\$

So, using this, we get

$\rm \: = \: \dfrac{1}{x} \times \displaystyle\lim_{h \to 0}\rm \frac{ {( \sqrt{x} )}^{2} - {( \sqrt{x + h}) }^{2} }{h( \sqrt{x} + \sqrt{x + h})}$

$\rm \: = \: \dfrac{1}{x} \times \displaystyle\lim_{h \to 0}\rm \frac{x - (x + h)}{h( \sqrt{x} + \sqrt{x + h})}$

$\rm \: = \: \dfrac{1}{x} \times \displaystyle\lim_{h \to 0}\rm \frac{x - x - h}{h( \sqrt{x} + \sqrt{x + h})}$

$\rm \: = \: \dfrac{1}{x} \times \displaystyle\lim_{h \to 0}\rm \frac{ - 1}{( \sqrt{x} + \sqrt{x + h})}$

$\rm \: = \: \dfrac{1}{x} \times \dfrac{ - 1}{( \sqrt{x} + \sqrt{x + 0})}$

$\rm \: = \: \dfrac{1}{x} \times \dfrac{ - 1}{ \sqrt{x} + \sqrt{x}}$

$\rm \: = \: \dfrac{1}{x} \times \dfrac{ - 1}{2\sqrt{x}}$

$\rm \: = \:\dfrac{ - 1}{2x\sqrt{x}}$

Hence,

$\rm\implies \:\boxed{\tt{ \frac{d}{dx} \frac{1}{ \sqrt{x} } \: = - \: \frac{1}{2x \sqrt{x} } \: }} \\$

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