Derivative of Cos 2x

What is the Derivative of Cos 2x ?

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

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

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

Proof of the Derivative of Cos 2x

Way 1

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

$(cos2x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{cos2(x+h)-cos2x}{h}$

$(cos2x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{cos(2x+2h)-cos2x}{h}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Way 2

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

$(cos2x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{cos2(x+h)-cos2x}{h}$

$(cos2x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{cos(2x+2h)-cos2x}{h}$

$\boxed{cosp-cosq=-2.sin\frac{p-q}{2}.sin\frac{p+q}{2}}$

$(cos2x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{-2.sin\frac{\cancel{2x}+2h-\cancel{2x}}{2}.sin\frac{2x+2h+2x}{2}}{h}$

$(cos2x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{-2.sin\frac{2h}{2}.sin\frac{4x+2h}{2}}{h}$

$(cos2x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{-2.sin\frac{\cancel{2}.h}{\cancel{2}}.sin\frac{\cancel{2}.(2x+h)}{\cancel{2}}}{h}$

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

$(cos2x{)}^{\prime }=\underset{h\to 0}{\lim }[\frac{-2.sinh}{h}.sin(2x+h)]$

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

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

$(cos2x{)}^{\prime }=-\mathrm{2.1.}sin(2x+0)$

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

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

Way 3

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

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

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

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

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

$(cos2x{)}^{\prime }=0-\frac{2.2x.(2x{)}^{\prime }}{2!}+\frac{4.(2x{)}^3.(2x{)}^{\prime }}{4!}-\frac{6.(2x{)}^5.(2x{)}^{\prime }}{6!}+\frac{8.(2x{)}^7.(2x{)}^{\prime }}{8!}-...$

$(cos2x{)}^{\prime }-\frac{2.2x.2}{2!}+\frac{4.(2x{)}^3.2}{4!}-\frac{6.(2x{)}^5.2}{6!}+\frac{8.(2x{)}^7.2}{8!}-...$

$(cos2x{)}^{\prime }=-\frac{\cancel{2}.2x.2}{\cancel{2}.1!}+\frac{\cancel{4}.(2x{)}^3.2}{\cancel{4}.3!}-\frac{\cancel{6}.(2x{)}^5.2}{\cancel{6}.5!}+\frac{\cancel{8}.(2x{)}^7.2}{\cancel{8}.7!}-...$

$(cos2x{)}^{\prime }=-\frac{2x.2}{1!}+\frac{(2x{)}^3.2}{3!}-\frac{(2x{)}^5.2}{5!}+\frac{(2x{)}^7.2}{7!}-...$

$(cos2x{)}^{\prime }=-\frac{2x.2}{1}+\frac{(2x{)}^3.2}{3!}-\frac{(2x{)}^5.2}{5!}+\frac{(2x{)}^7.2}{7!}-...$

$(cos2x{)}^{\prime }=-2x.2+\frac{(2x{)}^3.2}{3!}-\frac{(2x{)}^5.2}{5!}+\frac{(2x{)}^7.2}{7!}-...$

$(cos2x{)}^{\prime }=-2.2x+\frac{2.(2x{)}^3}{3!}-\frac{2.(2x{)}^5}{5!}+\frac{2.(2x{)}^7}{7!}-...$

$(cos2x{)}^{\prime }=-2.[2x-\frac{(2x{)}^3}{3!}+\frac{(2x{)}^5}{5!}-\frac{(2x{)}^7}{7!}+...]$

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