Question

find the derivative of the function 1/√x.
$\frac{1}{ \sqrt{x} }$
​

Answer 1

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

Given function is

$\rm :\longmapsto\:\dfrac{1}{ \sqrt{x} }$

Let assume that

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

can be further rewritten as

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

can be rewritten as

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

On differentiating both sides, w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}f(x) = \dfrac{d}{dx} {\bigg(x\bigg) }^{ \: - \: \dfrac{1}{2}}$

We know,

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1} \: }}}$

So, using this identity, we get

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

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

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

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

$\bf\implies \:f'(x) \: = \: - \: \dfrac{1}{2x \sqrt{x} }$

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More to know :-

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