Question

Differentiate the following function with respect to x using first principal

$f(x) = sin \sqrt{x}$
​

Answer 1

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

Given function is

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

So,

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

Using Definition of First Principal, we get

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

On substituting the values, we get

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

We know,

$\boxed{ \tt{ \: sinx - siny = 2cos\bigg[\dfrac{x + y}{2} \bigg]sin\bigg[\dfrac{x - y}{2} \bigg]}}$

So, using this identity, we get

$\rm \: = \: \displaystyle\lim_{h \to 0}\sf \frac{2cos\bigg[\dfrac{ \sqrt{x + h} + \sqrt{x} }{2} \bigg]sin\bigg[\dfrac{ \sqrt{x + h} - \sqrt{x} }{2} \bigg]}{h}$

$\rm \: = \: 2cos\bigg[\dfrac{ \sqrt{x} + \sqrt{x} }{2} \bigg]\displaystyle\lim_{h \to 0}\sf \frac{sin\bigg[\dfrac{ \sqrt{x + h} - \sqrt{x} }{2} \bigg]}{h}$

$\rm \: = \: 2cos\bigg[\dfrac{2 \sqrt{x}}{2} \bigg]\displaystyle\lim_{h \to 0}\sf \frac{sin\bigg[\dfrac{ \sqrt{x + h} - \sqrt{x} }{2} \bigg]}{\dfrac{ \sqrt{x + h} - \sqrt{x} }{2} } \times \dfrac{ \sqrt{x + h} - \sqrt{x} }{2h}$

We know,

$\ \red{\rm :\longmapsto\: \boxed{ \tt{\displaystyle\lim_{x \to 0}\sf \: \frac{sinx}{x} = 1 \: }}}$

$\rm \: = \: cos \sqrt{x} \: \displaystyle\lim_{h \to 0}\sf \dfrac{ \sqrt{x + h} - \sqrt{x} }{h}$

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

$\rm \: = \: cos \sqrt{x} \: \displaystyle\lim_{h \to 0}\sf \dfrac{x + h - x }{h \: [ \sqrt{x + h} + \sqrt{x} ]}$

$\rm \: = \: cos \sqrt{x} \: \displaystyle\lim_{h \to 0}\sf \dfrac{h}{h \: [ \sqrt{x + h} + \sqrt{x} ]}$

$\rm \: = \: cos \sqrt{x} \: \displaystyle\lim_{h \to 0}\sf \dfrac{1}{\: [ \sqrt{x + h} + \sqrt{x} ] \: }$

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

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

Hence,

$\rm \implies\:\boxed{ \tt{ \: \dfrac{d}{dx}sin \sqrt{x} = \frac{cos \sqrt{x} }{2 \sqrt{x} } \: }}$

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Additional Information :-

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

Answer 2

Question:-

Differentiate the following function with respect to x using first principal

$\sf \large \: f(x) = sin \sqrt{x}.$

Given:-

$\sf \large \: f(x) = sin \sqrt{x} .$

Solution:-

$\sf \large Then, f'(x) = \sf\tt{\displaystyle \sf \large \lim_{h \to 0} } \bigg[ \sf \frac{f(x + h) - f(x)}{h} \bigg ]$

$\sf \large= \sf\tt{\displaystyle \sf \lim_{h \to 0} } \bigg[ \sf\frac{sin( \sqrt{x + h} ) - sin( \sqrt{x}) }{h} \bigg]$

$\sf \large = \sf\tt{\displaystyle \sf \lim_{h \to 0} } \bigg[ \sf \large\frac{2 \: cos( \frac{ \sqrt{x + h} + \sqrt{x} }{2} )sin( \frac{ \sqrt{x + h} - \sqrt{x} }{2}) }{h} \bigg]$

$\sf \large = \sf\tt{\displaystyle \sf \lim_{h \to 0} } \bigg[ \sf \large2\frac{( \frac{ \sqrt{x + h - \sqrt{x} }) }{2})cos( \frac{ \sqrt{x + h} + \sqrt{x} }{2})sin( \frac{ \sqrt{x + h} - \sqrt{x} }{2} ) }{h( \frac{ \sqrt{x + h} - \sqrt{x} }{2} )} \bigg]$

$\sf \large = \sf\tt{\displaystyle \sf \lim_{h \to 0} } \bigg[ \sf \large\frac{ \sqrt{x +h } - \sqrt{x} }{h}\bigg ]\sf\tt{\displaystyle \sf \lim_{h \to 0} } \bigg[ \sf \large cos( \sf \large \frac{ \sqrt{x + h} + \sqrt{x} }{2}) \bigg ] \sf \large\sf\tt{\displaystyle \sf \lim_{h \to 0} } \bigg[ \sf \large\frac{ sin(\frac{ \sqrt{x + h} - \sqrt{x} }{2} )}{ \frac{ \sqrt{x + h} - \sqrt{x} }{2} } \bigg]$

$\sf \large = \sf\tt{\displaystyle \sf \lim_{x \to 0} } \bigg[ \sf \large\frac{(x + h) - x}{( \sqrt{x + h} + \sqrt{x} } \bigg ][cos \sqrt{x} \: ] \sf[1]$

$\sf \large = cos \sqrt{x} \: \: \sf\tt{\displaystyle \sf \lim_{x \to 0} } \bigg[ \sf \large\frac{1}{ \sqrt{x + h} + \sqrt{x} } \bigg]$

$\sf \large = cos \sqrt{x} \bigg [ \frac{1}{2 \sqrt{x} } \bigg] = \frac{cos \sqrt{x} }{2x}$

Answer:-

${ \boxed{ \sf \large \red{ \frac{d}{dx} sin \sqrt{x} = \frac{cos \sqrt{x} }{2x}. }}}$

Hope you have satisfied. ⚘