Question

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Answer 1

Step-by-step explanation:

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Answer 2

Given Question :-

Differentiate using first principle sqrt(x).

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

Let assume that

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

So,

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

Using Definition of First Principle, we have

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

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

On rationalizing the numerator, we get

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

We know,

$\boxed{ \bf{ \: (x + y)(x - y) = {x}^{2} - {y}^{2}}}$

So, using this,

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

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

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

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

$\rm :\longmapsto\:f'(x) = \dfrac{1}{\sqrt{x + 0} + \sqrt{x}}$

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

$\rm :\longmapsto\:f'(x) = \dfrac{1}{ \: \: 2\sqrt{x} \: \: }$

Hence,

$\boxed{ \qquad \bf{ \: \dfrac{d}{dx} \sqrt{x} = \frac{1}{2 \sqrt{x} } \qquad}}$

Additional Information :-

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