Question

Demonstre, pela definição de derivada, que se f(x) = senx, então f′(x) = con x

Answer 1

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

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

aplicando isso

$f'(x)=\lim_{h \to 0} \frac{2*cos( \frac{x+h+x}{2} )*sen( \frac{x+h-x}{2}) }{h}\\\\ \boxed{\boxed{f'(x)=\lim_{h \to 0} \frac{2*cos( \frac{2x+h}{2} )*sen( \frac{h}{2}) }{h}}}$

lembrando que: 
$\boxed{\boxed{ \lim_{x \to 0} \frac{sen(x)}{x}=1 }}$

dividindo o numerador e o denominador por 2

$f'(x)=\lim_{h \to 0} \frac{2*cos( \frac{2x+h}{2} )*sen( \frac{h}{2}) }{h}}} * \frac{ (\frac{1}{2} )}{ (\frac{1}{2}) } \\\\ f'(x)=\lim_{h \to 0} \frac{cos( \frac{2x+h}{2} )*sen( \frac{h}{2}) }{ \frac{h}{2} }}}\\\\\\ f'(x)=\lim_{h \to 0} cos( \frac{2x+h}{2} )* \frac{sen( \frac{h}{2} )}{\frac{h}{2}} \\\\ f'(x)= cos( \frac{2x+0}{2} ) * 1\\\\f'(x)=cos(x)$