Utilizando a definição, mostre que $(\sin x)' = \cos x$.

Utilizando a definição, mostre que $(\sin x)' = \cos x$.

Demonstração:

$(\sin x)' = \displaystyle\lim_{h \rightarrow 0} \dfrac{\sin (x+h) - \sin x}{h} = \displaystyle\lim_{h \rightarrow 0} \dfrac{(\sin x)(\cos h) + (\sin h)(\cos x) - \sin x}{h} =$

$= \displaystyle\lim_{h \rightarrow 0} \dfrac{(\sin x)(\cos h) - \sin x}{h} + \displaystyle\lim_{h \rightarrow 0} \dfrac{(\sin h)(\cos x)}{h} = \displaystyle\lim_{h \rightarrow 0} \dfrac{(\sin x)[(\cos h) - 1]}{h} + \displaystyle\lim_{h \rightarrow 0} \dfrac{(\sin h)(\cos x)}{h} =$

$= -\displaystyle\lim_{h \rightarrow 0} [(\sin x) \cdot \dfrac{(\sin \dfrac{h}{2})}{h/2} \cdot (\sin \dfrac{h}{2})] + \displaystyle\lim_{h \rightarrow 0} \dfrac{(\sin h)(\cos x)}{h} =$

$= \cancelto{0}{-(\displaystyle\lim_{h \rightarrow 0} \sin x) \cdot \underset{1}{\underbrace{(\displaystyle\lim_{h \rightarrow 0} \dfrac{(\sin \dfrac{h}{2})}{h/2})}} \cdot (\displaystyle\lim_{h \rightarrow 0} \sin \dfrac{h}{2})} + \displaystyle\lim_{h \rightarrow 0} \dfrac{(\sin h)(\cos x)}{h} =$

$= \underset{1}{\underbrace{\displaystyle\lim_{h \rightarrow 0} \dfrac{\sin h}{h}}} \cdot \cos x$

Logo $\fbox{$(\sin x)' = \cos x$}$.