Question

usando as regras de derivação,calcule as derivadas das funçoes:

a)-f(x)= 3x2 senx

b)- f(x)= 3 senx.log 3 x
c)-f(x) = sec x
d)- f(x)= 2x+1
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X2 +3

Answer 1

a) usando a regra do produto  (U.V) = U'.V + U.V' 
U = 3x² 
U' = 6x
V = sen(x)
V' = cos(x) 
$f'(x)=6x*sen(x)+3x^2*cos(x)\\\\\text{simplificando }\\\\f'(x)=3x*[2sen(x)+xcos(x)]$

b) regra do produto novamente
U = 3sen(x)
U' =3cos(x)
V= log(3x)
V' = 1/(3x) * 3 = 1/x

$f'(x)= 3cos(x)*log(x) +3sen(x)* \frac{1}{x} \\\\f'(x)=3*[ cos(x)log(x)+ \frac{sen(x)}{x} ]\\\\f'(x)=3*[ \frac{3cos(x)log(x)+sen(x)}{3}]$

c) se vc olhar na tabela a derivada de sec(u) = u'.sec(u).tg(u)
então a sua derivada ficaria f'(x)=sec(x).tg(x)

se vc quiser calcular 
$f(x)=sec(x)\\\\f'(x)= \frac{1}{cos(x)} = (cos(x))^{-1}$

derivando
$f'(x) = -1*(cos(x))^{-2}*(-sen(x))\\\\f'(x)= \frac{sen(x)}{cos^2(x)} \\\\f'(x)= \frac{1}{cos(x)}* \frac{sen(x)}{cos(x)} \\\\f'(x)=sec(x)*tg(x)$

d) usando a regra do quociente 
(U/V)' = [ U'.V -U.V']/V²

U=2x+1
U'=2
V=x²+3
V' =2x


$f'(x)= \frac{2*(x^2+3) -(2x+1)*(2x)}{(x^2+3)^2} \\\\\f'(x)= \frac{2x^2+6 -(4x^2+2x)}{(x^2+3)^2} \\\\f'(x)= \frac{2x^2+6-4x^2-2x}{(x^2+3)^2} \\\\f'(x)= \frac{-2x^2-2x+6}{(x^2+3)^2} = \frac{2(-x^2-2x+3)}{(x^2+3)^2}$