Question

5. Calcule f0(x), usando a definição.
a) f(x) = cos(x).
b) f(x) = 20.

Answer 1

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

aplicando isso 
a)
$\lim_{h \to 0} \frac{cos(x+h)-cos(x)}{h} \\\\ \lim_{h \to 0} \frac{cos(x)*cos(h)-sen(x)*sen(h)-cos(x)}{h} \\\\ \lim_{h \to 0} \frac{cos(x)*[cos(h)-1]-sen(x)*sen(h)}{h} \\\\ \boxed{\boxed{ \lim_{h \to 0} \frac{cos(x)*[cos(h)-1]}{h} - \lim_{h \to 0} \frac{sen(x)*sen(h)}{h} }}$

limite fundamental
lim x->0  sen(x)/x = 1

ficando
$\boxed{\boxed{ \lim_{h \to 0} \frac{cos(x)*[cos(h)-1]}{h} - sen(x)*1 }}$

resolvendo o outro limite
multiplicando pelo conjugado

$\lim_{h \to 0} \frac{cos(x)*[cos(h)-1]}{h} \\\\ cos(x)*\lim_{h \to 0} \frac{[cos(h)-1]}{h} \\\\ cos(x)*\lim_{h \to 0} \frac{[cos(h)-1]}{h} * \frac{[cos(h)+1]}{[cos(h)+1]} \\\\ cos(x)*\lim_{h \to 0} \frac{cos^2(h)-1^2}{h*[cos(h)+1]} \\\\ cos(x)*\lim_{h \to 0} \frac{sen^2(h)}{h*[cos(h)+1]}\\\\ cos(x)*\lim_{h \to 0} \frac{sen(h)}{h}* \frac{sen(h)}{[cos(h)+1]} \\\\ cos(x)*1* \frac{sen(0)}{cos(0)+1} =0 \\\\\.$

então
$f'(x)=\lim_{h \to 0} \frac{cos(x)*[cos(h)-1]}{h} - sen(x)*1 \\\\f'(x)=0-sen(x)\\\\f'(x)=-sen(x)$

b)
$f'(x)= \lim_{h \to 0} \frac{20-20}{h} \\\\f'(x)= \lim_{h \to 0} \frac{0}{h} \\\\f'(x)=0$