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Differentiate with respect to x using first principal

$f(x) = {e}^{sin2x}$

Class - 11

Topic - Differentiation ​

Answer 1

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

Given function is

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

So,

$\rm :\longmapsto\:f(x + h) = {e}^{sin2(x + h)} = {e}^{sin(2x + 2h)}$

By using definition of First Principal, we have

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

So, on substituting the values, we get

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

$\rm \: = \:\displaystyle\lim_{h \to 0} \frac{{e}^{sin2x}\bigg[{e}^{sin(2x + 2h) - sin2x} - 1\bigg]}{h}$

$\rm \: = \:\displaystyle\lim_{h \to 0} \frac{{e}^{sin2x}\bigg[{e}^{sin(2x + 2h) - sin2x} - 1\bigg]}{[sin(2x + 2h) - sin2x]h} \times [sin(2x + 2h) - sin2x]$

We know,

$\boxed{ \tt{ \: \displaystyle\lim_{x \to 0} \frac{ {e}^{x} - 1}{x} = 1 \: }}$

So, using this identity, we get

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

We know,

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

So, using this, we have

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

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

$\rm \: = \:{e}^{sin2x}\displaystyle\lim_{h \to 0} \frac{2cos\bigg[\dfrac{4x + 2h}{2} \bigg]sinh}{h}$

$\rm \: = \:2{e}^{sin2x}\displaystyle\lim_{h \to 0}cos\bigg[\dfrac{4x + 2h}{2} \bigg] \times \displaystyle\lim_{h \to 0} \frac{sinh}{h}$

$\rm \: = \:2{e}^{sin2x} \times cos2x \times 1$

$\rm \: = \:2cos2x \: {e}^{sin2x}$

Hence,

$\bf\implies \:\boxed{ \tt{ \: \dfrac{d}{dx}{e}^{sin2x} = 2cos2x \: {e}^{sin2x} \: }}$

More to know :-

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