Question

encontre f' (1), sendo f(x)=(6x+1)(2x+4)​

Answer 1

Podemos fazer de duas formas  :

1ª forma : Pela regra do produto

$\text{f . g} = \text{f' . g + f . g' } \\\\ \text{Ent{\^a}o}\\\\ \text{f(x)} = (6\text x+1)(2\text x+4) \\\\ \text{f'(x)} = \text{(6x+1)'.(2\text x+4) + (6x+1)(2x +4)'} \\\\ \text{f'(x)}=6(2\text x+4) + (6\text x+1).2 \\\\ \text{f'(x)}=12\text x+24+12\text x+2 \\\\ \text{f'(x)}=24\text x+26 \\\\ \text{Fa{\c c}amos x = 1}\\\\ \text{f'(1)} = 24+26 \\\\ \huge\boxed{\text{f'(1)} = 50}\checkmark$

2ª forma

Faz a distributiva e deriva normalmente :

$\text{f(x)}=\text{(6x+1)(2x+4)} \\\\ \text{f'(x)} = 12\text x^2+24\text x+2\text x+4 \\\\ \text{f(x)}=12\text x^2+26\text x+4 \\\\ \text{Derivando} :\\\\ \text{f'(x)} = 24\text x+26 \\\\ \text{x = 1} : \\\\ \text{f'(1)} = 24+26 \\\\ \huge\boxed{\text{f'(1)} = 50}\checkmark$