Question

QUESTÃO 5
Dadas as funções:

Assim, z(x) = f’(x)+g’(x)+h’(x) é igual a:

Answer 1

$\displaystyle \sf f(x) = 3x^2-2cos(x) \\\\ g(x) = x^3\cdot cos(x) \\\\ h(x) = \frac{x\cdot sen(X)}{1+x} \\\\\\ z(x) = f'(x)+g'(x)+h'(x) \\\\ {Derivando} : \\\\ f'(x) = (3x^2-2cos(x))' \to f'(x) = (3x^2)'-2cos'(x) \\\\ f'(x) =6x - 2\cdot (-sen(x)) \\\\\ f'(x) = 6x+2sen(x)$

$\displaystyle \sf g(x) = x^3\cdot cos(x) \\\\ \text{Aplique a regra do produto }:\\\\\ [f\cdot g]'=f'\cdot g+f\cdot g' \\\\ Da{\'i}}: \\\\ g'(x) = (x^3)'\cdot cos(x) +x^3\cdot (cos(x))' \\\\ g'(x) =3x^2\cdot cos(x) -x^3\cdot sen(x)$

$\displaystyle \sf h(x) = \frac{x\cdot sen(x) }{1+x} \\\\ \text{Aplique a regra do quociente } : \\\\ \left(\frac{f}{g}\right)' = \frac{f'\cdot g-f\cdot g'}{g^2 } \\\\\\ Da{\'i}}: \\\\ h'(x) = \frac{(x\cdot sen(x))'\cdot (1+x) - (x\cdot sen(x))\cdot (1+x)' }{(1+x)^2}$

$\displaystyle \sf h'(x) = \frac{[x'\cdot sen(x)+x\cdot sen'(x)]\cdot (1+x)-x\cdot sen(x) \cdot 1 }{(1+x)^2 } \\\\\\\ h'(x) = \frac{[sen(x)+x\cdot cos(x)](1+x)-xsen(x) }{(1+x)^2} \\\\\\ h'(x) =\frac{sen(x)+x\cdot sen(x)+x\cdot cos(x)+x^2\cdot cos(x)-x\cdot sen(x) }{(1+x)^2} \\\\\\ h'(x) = \frac{(x^2+x)\cdot cos(x)+sen(x) }{(1+x)^2}$

Daí :

$\displaystyle \sf \boxed{\sf \ z(x) = 6x+2sen(x)+3x^2cos(X)-x^3sen(x)+\frac{(x^2+x)cos(x)+sen(x)}{(1+x)^2} \ }\checkmark$

última alternativa