Производная от Sin x

Какова Производная от Sin x?

Производная от sin x равна cos x.

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

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

Доказательство Производной от Sin x

Метод 1

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

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

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

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

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

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

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

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

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

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

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

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

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

Метод 2

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

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

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

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

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

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

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

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

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

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

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

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

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

$(sinx{)}^{\prime }=1.cos(x+0)$

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

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

Метод 3

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

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

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

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

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

$(sinx{)}^{\prime }=1-\frac{3x^2}{3!}+\frac{5x^4}{5!}-\frac{7x^6}{7!}+\frac{9x^8}{9!}-...$

$(sinx{)}^{\prime }=1-\frac{\cancel{3}{x}^2}{\cancel{3}.2!}+\frac{\cancel{5}{x}^4}{\cancel{5}.4!}-\frac{\cancel{7}{x}^6}{\cancel{7}.6!}+\frac{\cancel{9}{x}^8}{\cancel{9}.8!}-...$

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

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

Метод 4

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

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

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

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

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

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

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

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

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

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

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

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

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

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