Производная от Cos x

Какова Производная от Sin x?

Производная от cos x равна -sin x.

$(cosx{)}^{\prime }=-sinx$

$\frac{d}{dx}(cosx)=-sinx$

Доказательство Производной от Cos x

Метод 1

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

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

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

$(cosx{)}^{\prime }=\underset{h\to 0}{\lim }\frac{cosx.cosh-sinx.sinh-cosx}{h}$

$(cosx{)}^{\prime }=\underset{h\to 0}{\lim }\frac{cosx.cosh-cosx-sinx.sinh}{h}$

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

$(cosx{)}^{\prime }=\underset{h\to 0}{\lim }[\frac{cosx.(cosh-1)}{h}-\frac{sinx.sinh}{h}]$

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

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

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

$(cosx{)}^{\prime }=cosx.0-sinx.1$

$(cosx{)}^{\prime }=0-sinx$

$(cosx{)}^{\prime }=-sinx$

Метод 2

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

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

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

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

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

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

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

$(cosx{)}^{\prime }=\underset{h\to 0}{\lim }\frac{-sin\frac{h}{2}.sin(x+\frac{h}{2})}{\frac{1}{2}.h}$

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

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

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

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

$h\to 0(\frac{h}{2}=h)$

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

$(cosx{)}^{\prime }=-1.sin(x+0)$

$(cosx{)}^{\prime }=-1.sinx$

$(cosx{)}^{\prime }=-sinx$

Метод 3

$\boxed{sinx=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-...}$

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

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

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

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

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

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

$(cosx{)}^{\prime }=-\frac{x}{1!}+\frac{x^3}{3!}-\frac{x^5}{5!}+\frac{x^7}{7!}-...$

$(cosx{)}^{\prime }=-\frac{x}{1}+\frac{x^3}{3!}-\frac{x^5}{5!}+\frac{x^7}{7!}-...$

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

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

$(cosx{)}^{\prime }=-sinx$

Метод 4

$\boxed{e^{ix}=cosx+i.sinx}$

$(cosx+i.sinx{)}^{\prime }=(e^{ix}{)}^{\prime }$

$(cosx{)}^{\prime }+(i.sinx{)}^{\prime }=(e^{ix}{)}^{\prime }$

$(cosx{)}^{\prime }+i.(sinx{)}^{\prime }=(e^{ix}{)}^{\prime }$

$\boxed{(e^u{)}^{\prime }=u^{\prime }.e^u}$

$(cosx{)}^{\prime }+i.(sinx{)}^{\prime }=(ix{)}^{\prime }.e^{ix}$

$(cosx{)}^{\prime }+i.(sinx{)}^{\prime }=i.e^{ix}$

$(cosx{)}^{\prime }+i.(sinx{)}^{\prime }=i.(cosx+i.sinx)$

$(cosx{)}^{\prime }+i.(sinx{)}^{\prime }=i.cosx+i^2.sinx$

$\boxed{i^2=-1}$

$(cosx{)}^{\prime }+i.(sinx{)}^{\prime }=i.cosx+(-1).sinx$

$(cosx{)}^{\prime }+i.(sinx{)}^{\prime }=i.cosx+(-sinx)$

$(cosx{)}^{\prime }+i.(sinx{)}^{\prime }=-sinx+i.cosx$

$(cosx{)}^{\prime }=-sinx$