Question

Derivada de sen x pela definição

Answer 1

$f(x)=sen(x)\\\\\boxed{\boxed{f'(x)= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} }}$

então temos
$\lim_{h \to 0} \frac{sen(x+h)-sen(x)}{h}$

lembrando que 
$\boxed{\boxed{sen(a)-sen(b)= 2*sen( \frac{a-b}{2} )*cos( \frac{a+b}{2} )}}$

neste caso
a =(x+h)
b = x

ficando com 
$\lim_{h \to 0} \frac{2*sen( \frac{x+h-x}{2} )*cos( \frac{x+x+h}{2} )}{h}\\\\\lim_{h \to 0} \frac{2*sen( \frac{h}{2} )*cos( \frac{2x+h}{2} )}{h}\\\\ \boxed{\lim_{h \to 0}\frac{2*sen( \frac{h}{2} )*cos( x+\frac{h}{2} )}{h}}$

aplicando uma mudança de variavel
fazendo
$\boxed{u= \frac{h}{2} }$
então
$\boxed{2*u =h}$

quando h tende a 0  ...u tambem tende a 0 
porque vc teria
$u= \frac{0}{2}=0$

então podemos reescrever o limite como
$\lim_{u \to 0} \frac{\not 2*sen(u)*cos(x+u)}{\not 2u} \\\\ \lim_{u \to 0} \frac{sen(u)*cos(x+u)}{u} \\\\\ \boxed{\lim_{u \to 0} \frac{sen(u)}{u}*cos(x+u) }$

pelo limite fundamental sabemos que 
$\boxed{\boxed{ \lim_{x \to 0} \frac{sen(x)}{x} =1}}$

calculando o limite
$\lim_{u \to 0} \frac{sen(u)}{u}*cos(x+u) }= 1*cos(x+0) = cos(x)$

...

$\boxed{\boxed{f'(x)= \lim_{h \to 0} \frac{sen(x+h)-sen(x)}{h} =cos(x)}}$
.

Answer 2

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Limites trigonométricos fundamentais.

$\Large\boxed{\boxed{\boxed{\boxed{\displaystyle\sf\lim_{x \to 0}\dfrac{sen(x)}{x}=1}}}}\\\boxed{\boxed{\boxed{\boxed{\displaystyle\sf\lim_{x \to 0}\dfrac{1-cos(x)}{x}=0}}}}$

Definição de derivada

$\Large\boxed{\boxed{\boxed{\boxed{\displaystyle\sf f'(x)=\lim_{h \to 0}\dfrac{f(x+h)-f(x)}{h}}}}}$

$\sf f(x)=sen(x)\\\displaystyle\sf f'(x)=\lim_{h \to 0}\dfrac{f(x+h)-f(x)}{h}\\\displaystyle\sf f'(x)=\lim_{h \to 0}\dfrac{sen(x+h)-sen(x)}{h}\\\displaystyle\sf f'(x)=\lim_{h \to 0}\dfrac{sen(x)\cdot cos(h)+sen(h)\cdot cos(x)-sen(x)}{h}\\\displaystyle\sf f'(x)=\lim_{h \to 0}\dfrac{sen(x)\cdot cos(h)-sen(x)+sen(h)\cdot cos(x)}{h}\\\displaystyle\sf f'(x)=\lim_{h \to 0}\dfrac{sen(x)\cdot cos(h)-sen(x)}{h}+\lim_{h \to 0}\dfrac{sen(h)\cdot cos(x)}{h}$

$\displaystyle\sf f'(x)=\lim_{h \to 0}\dfrac{sen(x)[cos(h)-1]}{h}+\lim_{h \to 0}\dfrac{sen(h)}{h}\cdot\lim_{h \to 0}cos(x)\\\displaystyle\sf f'(x)= \lim_{h \to 0}sen(x)\cdot\lim_{h \to 0}\dfrac{cos(h)-1}{h}+\lim_{h \to 0}\dfrac{sen(h)}{h}\cdot\lim_{h \to 0}cos(x)\\\sf f'(x)=\lim_{h \to 0}sen(x)\cdot-1\cdot\lim_{h \to 0}\dfrac{1-cos(h)}{h}+\lim_{h \to 0}\dfrac{sen(h)}{h}\cdot\lim_{h \to 0}cos(x)\\\displaystyle\sf f'(x)=sen(x)\cdot-1\cdot 0+1\cdot cos(x)\\\huge\boxed{\boxed{\boxed{\boxed{\sf f'(x)=cos(x)}}}}$