Question

Differentiate using first principal

$f(x) = \sqrt{sin2x}$

​

Answer 1

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

Given function is

$\rm :\longmapsto\:f(x) = \sqrt{sin2x} \:$

So,

$\rm :\longmapsto\:f(x + h) = \sqrt{sin2x + 2h} \:$

By Using, Definition of First Principal, we have

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

So, on substituting the values, we get

$\rm :\longmapsto\:f'(x) = \displaystyle\lim_{h \to 0} \frac{ \sqrt{sin(2x + 2h)} - \sqrt{sin2x} }{h}$

On rationalizing the denominator, we get

$\rm=\displaystyle\lim_{h \to 0} \: \frac{ \sqrt{sin(2x + 2h)} - \sqrt{sin2x} }{h} \times \frac{\sqrt{sin(2x + 2h)} + \sqrt{sin2x}}{\sqrt{sin(2x + 2h)} + \sqrt{sin2x}}$

We know that

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

So, using this, we get

$\rm \: = \: \displaystyle\lim_{h \to 0} \frac{sin(2x + 2h) - sin2x}{h(\sqrt{sin(2x + 2h)} + \sqrt{sin2x})}$

$\rm \:  =  \:\displaystyle\lim_{h \to 0} \frac{1}{\sqrt{sin(2x + 2h)} + \sqrt{sin2x}} \times \displaystyle\lim_{h \to 0} \frac{sin(2x + 2h) - sin2x}{h}$

$\rm \:=\dfrac{1}{\sqrt{sin2x} + \sqrt{sin2x}} \times \displaystyle\lim_{h \to 0} \frac{sin(2x + 2h) - sin2x}{h}$

We know that

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

So, using this identity, we get

$\rm \:  =  \:\dfrac{1}{2 \sqrt{sin2x} } \displaystyle\lim_{h \to 0} \frac{2cos\bigg[\dfrac{2x + 2h + 2x}{2} \bigg]sin\bigg[\dfrac{2x + 2h - 2x}{2} \bigg]}{h}$

$\rm \:  =  \:\dfrac{1}{2 \sqrt{sin2x} } \displaystyle\lim_{h \to 0} \frac{2cos\bigg[\dfrac{4x + 2h}{2} \bigg]sin\bigg[\dfrac{2h}{2} \bigg]}{h}$

$\rm \:  =  \:\dfrac{1}{2 \sqrt{sin2x} } \displaystyle\lim_{h \to 0} \frac{2cos\bigg[\dfrac{4x + 2h}{2} \bigg]sin\bigg[h \bigg]}{h}$

We know,

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

So, using this, we get

$\rm \:  =  \:\dfrac{1}{2 \sqrt{sin2x} } \times 2cos\bigg[\dfrac{4x}{2} \bigg] \times 1$

$\rm \:  =  \:\dfrac{1}{\sqrt{sin2x} } \times cos\bigg[2x\bigg]$

$\rm \:  =  \:\dfrac{cos2x}{ \sqrt{sin2x} }$

Hence,

$\rm \:\rm \implies\:\boxed{ \tt{ \: \frac{d}{dx} \sqrt{sin2x}   =  \:\dfrac{cos2x}{ \sqrt{sin2x} } \: }}$

More to know :-

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

Answer 2

Answer:

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

Given function is

$\rm :\longmapsto\:f(x) = \sqrt{sin2x} \:$

So,

$\rm :\longmapsto\:f(x + h) = \sqrt{sin2x + 2h} \:$

By Using, Definition of First Principal, we have

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

So, on substituting the values, we get

$\rm :\longmapsto\:f'(x) = \displaystyle\lim_{h \to 0} \frac{ \sqrt{sin(2x + 2h)} - \sqrt{sin2x} }{h}$

On rationalizing the denominator, we get

$\rm=\displaystyle\lim_{h \to 0} \: \frac{ \sqrt{sin(2x + 2h)} - \sqrt{sin2x} }{h} \times \frac{\sqrt{sin(2x + 2h)} + \sqrt{sin2x}}{\sqrt{sin(2x + 2h)} + \sqrt{sin2x}}$

We know that

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

So, using this, we get

$\rm \: = \: \displaystyle\lim_{h \to 0} \frac{sin(2x + 2h) - sin2x}{h(\sqrt{sin(2x + 2h)} + \sqrt{sin2x})}$

$\rm \:  =  \:\displaystyle\lim_{h \to 0} \frac{1}{\sqrt{sin(2x + 2h)} + \sqrt{sin2x}} \times \displaystyle\lim_{h \to 0} \frac{sin(2x + 2h) - sin2x}{h}$

$\rm \:=\dfrac{1}{\sqrt{sin2x} + \sqrt{sin2x}} \times \displaystyle\lim_{h \to 0} \frac{sin(2x + 2h) - sin2x}{h}$

We know that

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

So, using this identity, we get

$\rm \:  =  \:\dfrac{1}{2 \sqrt{sin2x} } \displaystyle\lim_{h \to 0} \frac{2cos\bigg[\dfrac{2x + 2h + 2x}{2} \bigg]sin\bigg[\dfrac{2x + 2h - 2x}{2} \bigg]}{h}$

$\rm \:  =  \:\dfrac{1}{2 \sqrt{sin2x} } \displaystyle\lim_{h \to 0} \frac{2cos\bigg[\dfrac{4x + 2h}{2} \bigg]sin\bigg[\dfrac{2h}{2} \bigg]}{h}$

$\rm \:  =  \:\dfrac{1}{2 \sqrt{sin2x} } \displaystyle\lim_{h \to 0} \frac{2cos\bigg[\dfrac{4x + 2h}{2} \bigg]sin\bigg[h \bigg]}{h}$

We know,

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

So, using this, we get

$\rm \:  =  \:\dfrac{1}{2 \sqrt{sin2x} } \times 2cos\bigg[\dfrac{4x}{2} \bigg] \times 1$

$\rm \:  =  \:\dfrac{1}{\sqrt{sin2x} } \times cos\bigg[2x\bigg]$

$\rm \:  =  \:\dfrac{cos2x}{ \sqrt{sin2x} }$

Hence,

$\rm \:\rm \implies\:\boxed{ \tt{ \: \frac{d}{dx} \sqrt{sin2x}   =  \:\dfrac{cos2x}{ \sqrt{sin2x} } \: }}$

More to know :-

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