Derivative of Sin 2x

What is the Derivative of Sin 2x ?

The derivative of sin 2x is 2.cos 2x.

$(sin2x{)}^{\prime }=2.cos2x$

$\frac{d}{dx}(sin2x)=2.cos2x$

Proof of the Derivative of Sin 2x

Way 1

$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}$

$(sin2x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{sin2(x+h)-sin2x}{h}$

$(sin2x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{sin(2x+2h)-sin2x}{h}$

$\boxed{sin(p+q)=sinp.cosq+cosp.sinq}$

$(sin2x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{sin2x.cos2h+cos2x.sin2h-sin2x}{h}$

$(sin2x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{sin2x.cos2h-sin2x+cos2x.sin2h}{h}$

$(sin2x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{sin2x.(cos2h-1)+cos2x.sin2h}{h}$

$(sin2x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{2.[sin2x.(cos2h-1)+cos2x.sin2h]}{2.h}$

$(sin2x{)}^{\prime }=2.\underset{h\to 0}{\lim }\frac{sin2x.(cos2h-1)+cos2x.sin2h}{2h}$

$(sin2x{)}^{\prime }=2.\underset{h\to 0}{\lim }[\frac{sin2x.(cos2h-1)}{2h}+\frac{cos2x.sin2h}{2h}]$

$(sin2x{)}^{\prime }=2.\underset{h\to 0}{\lim }\frac{sin2x.(cos2h-1)}{2h}+2.\underset{h\to 0}{\lim }\frac{cos2x.sin2h}{2h}$

$(sin2x{)}^{\prime }=2.sin2x.\underset{h\to 0}{\lim }\frac{cos2h-1}{2h}+2.cos2x.\underset{h\to 0}{\lim }\frac{sin2h}{2h}$

$h\to 0(2h=h)$

$(sin2x{)}^{\prime }=2.sin2x.\underset{h\to 0}{\lim }\frac{cosh-1}{h}+2.cos2x.\underset{h\to 0}{\lim }\frac{sinh}{h}$

$\boxed{\underset{t\to 0}{\lim }\frac{sint}{t}=1}$ $\boxed{\underset{t\to 0}{\lim }\frac{cost-1}{t}=0}$

$(sin2x{)}^{\prime }=2.sin2x.0+2.cos2x.1$

$(sin2x{)}^{\prime }=0+2.cos2x$

$(sin2x{)}^{\prime }=2.cos2x$

Way 2

$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}$

$(sin2x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{sin2(x+h)-sin2x}{h}$

$(sin2x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{sin(2x+2h)-sin2x}{h}$

$\boxed{sinp-sinq=2.sin\frac{p-q}{2}.cos\frac{p+q}{2}}$

$(sin2x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{2.sin\frac{\cancel{2x}+2h-\cancel{2x}}{2}.cos\frac{2x+2h+2x}{2}}{h}$

$(sin2x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{2.sin\frac{2h}{2}.cos\frac{4x+2h}{2}}{h}$

$(sin2x{)}^{\prime }=2.\underset{h\to 0}{\lim }\frac{sin\frac{2h}{2}.cos\frac{4x+2h}{2}}{h}$

$(sin2x{)}^{\prime }=2.\underset{h\to 0}{\lim }\frac{sin\frac{\cancel{2}.h}{\cancel{2}}.cos\frac{\cancel{2}.(2x+h)}{\cancel{2}}}{h}$

$(sin2x{)}^{\prime }=2.\underset{h\to 0}{\lim }\frac{sinh.cos(2x+h)}{h}$

$(sin2x{)}^{\prime }=2.\underset{h\to 0}{\lim }[\frac{sinh}{h}.cos(2x+h)]$

$(sin2x{)}^{\prime }=2.\underset{h\to 0}{\lim }\frac{sinh}{h}.\underset{h\to 0}{\lim }cos(2x+h)$

$(sin2x{)}^{\prime }=\mathrm{2.1.}cos(2x+0)$

$(sin2x{)}^{\prime }=\mathrm{2.1.}cos2x$

$(sin2x{)}^{\prime }=2.cos2x$

Way 3

$\boxed{sin2x=2x-\frac{(2x{)}^3}{3!}+\frac{(2x{)}^5}{5!}-\frac{(2x{)}^7}{7!}+\frac{(2x{)}^9}{9!}-...}$

$\boxed{cos2x=1-\frac{(2x{)}^2}{2!}+\frac{(2x{)}^4}{4!}-\frac{(2x{)}^6}{6!}+\frac{(2x{)}^8}{8!}-...}$

$sin2x=2x-\frac{(2x{)}^3}{3!}+\frac{(2x{)}^5}{5!}-\frac{(2x{)}^7}{7!}+\frac{(2x{)}^9}{9!}-...$

$(sin2x{)}^{\prime }=[2x-\frac{(2x{)}^3}{3!}+\frac{(2x{)}^5}{5!}-\frac{(2x{)}^7}{7!}+\frac{(2x{)}^9}{9!}-...{]}^{\prime }$

$(sin2x{)}^{\prime }=(2x{)}^{\prime }-[\frac{(2x{)}^3}{3!}{]}^{\prime }+[\frac{(2x{)}^5}{5!}{]}^{\prime }-[\frac{(2x{)}^7}{7!}{]}^{\prime }+[\frac{(2x{)}^9}{9!}{]}^{\prime }-...$

$(sin2x{)}^{\prime }=2-\frac{3.(2x{)}^2.(2x{)}^{\prime }}{3!}+\frac{5.(2x{)}^4.(2x{)}^{\prime }}{5!}-\frac{7.(2x{)}^6.(2x{)}^{\prime }}{7!}+\frac{9.(2x{)}^8.(2x{)}^{\prime }}{9!}-...$

$(sin2x{)}^{\prime }=2-\frac{3.(2x{)}^2.2}{3!}+\frac{5.(2x{)}^4.2}{5!}-\frac{7.(2x{)}^6.2}{7!}+\frac{9.(2x{)}^8.2}{9!}-...$

$(sin2x{)}^{\prime }=2-\frac{\cancel{3}.(2x{)}^2.2}{\cancel{3}.2!}+\frac{\cancel{5}.(2x{)}^4.2}{\cancel{5}.4!}-\frac{\cancel{7}.(2x{)}^6.2}{\cancel{7}.6!}+\frac{\cancel{9}.(2x{)}^8.2}{\cancel{9}.8!}-...$

$(sin2x{)}^{\prime }=2-\frac{(2x{)}^2.2}{2!}+\frac{(2x{)}^4.2}{4!}-\frac{(2x{)}^6.2}{6!}+\frac{(2x{)}^8.2}{8!}-...$

$(sin2x{)}^{\prime }=2.[1-\frac{(2x{)}^2}{2!}+\frac{(2x{)}^4}{4!}-\frac{(2x{)}^6}{6!}+\frac{(2x{)}^8}{8!}-...]$

$(sin2x{)}^{\prime }=2.cos2x$