Question

if y=tan2x,find dy/dx from defination

Answer 1

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

Given function is

$\rm :\longmapsto\:y = \: tan2x \:$

Let assume that

$\rm :\longmapsto\:f(x) = tan2x$

So,

$\rm :\longmapsto\:f(x + h) = tan2(x +h) = tan(2x + 2h)$

By definition, we have

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

On substituting the values, we get

$\rm \: = \:\displaystyle\lim_{h \to 0} \frac{tan(2x + 2h) - tan2x}{h}$

can be rewritten as

$\rm \: = \:\displaystyle\lim_{h \to 0} \frac{1}{h}\bigg[\dfrac{sin(2x + 2h)}{cos(2x + 2h)} - \dfrac{sin2x}{cos2x} \bigg]$

$\rm \: = \:\displaystyle\lim_{h \to 0} \frac{1}{h}\bigg[\dfrac{sin(2x + 2h)cos2x - sin2xcos(2x + 2h)}{cos(2x + 2h)cos2x} \bigg]$

We know,

$\boxed{ \tt{ \: sinxcosy + sinycosx = sin(x + y) \: }}$

So, using this, we get

$\rm \: = \:\displaystyle\lim_{h \to 0} \frac{1}{h}\bigg[\dfrac{sin(2x + 2h - 2x)}{cos(2x + 2h)cos2x} \bigg]$

$\rm \: = \:\displaystyle\lim_{h \to 0} \frac{1}{h}\bigg[\dfrac{sin2h}{cos(2x + 2h)cos2x} \bigg]$

$\rm \: = \:\dfrac{1}{cos2x} \times \displaystyle\lim_{h \to 0} \frac{1}{cos(2x + 2h)} \times \displaystyle\lim_{h \to 0} \frac{sin2h}{h}$

$\rm \: = \:\dfrac{1}{cos2x} \times \dfrac{1}{cos2x} \times \displaystyle\lim_{h \to 0} \frac{sin2h}{2h} \times 2$

We know,

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

So, using this, we get

$\rm \: = \:\dfrac{1}{cos2x} \times \dfrac{1}{cos2x} \times \times 2$

$\rm \: = \:\dfrac{2}{ {cos}^{2}2x}$

$\rm \: = \: {2sec}^{2}2x$

Hence,

$\bf\implies \:\boxed{ \tt{ \: \frac{d}{dx} tan2x = 2 {sec}^{2}2x \: }}$

More to know :-

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

Answer 2

Answer:

$</p><p>\bf\implies \:\boxed{ \tt{ \: \frac{d}{dx} tan2x = 2 {sec}^{2}2x \: }}</p><p></p><p>$

Step-by-step explanation:

Hope it helps you