Question

URGENTE
determinar f'(0) recorrendo a definição de derivada de uma função
f(x)= x+sen(2x)​

Answer 1

 Derivada no ponto

$\displaystyle\huge\mathsf{f'(x_{0})=\lim_{x \to x_{0}}\dfrac{f(x)-f(x_{0})}{x-x_{0}}}$

$\mathsf{f(0)=0+sen(2.0)=0}$

$\displaystyle\mathsf{f'(0)=\lim_{x \to 0}\dfrac{x-sen(2x)-f(0)}{x-0}}$

$\displaystyle\mathsf{f'(0)=\lim_{x \to 0}\dfrac{x+sen(2x)}{x}}\\\displaystyle\mathsf{f'(0)=\lim_{x \to 0}\dfrac{x}{x}+\lim_{x \to 0}\dfrac{sen(2x)}{x}}$

Vamos resolver cada limite separadamente e depois somar.

$\displaystyle\mathsf{\lim_{x \to 0}\dfrac{x}{x}\to~\lim_{x \to 0}1=1}$

$\displaystyle\mathsf{\lim_{x \to 0}\dfrac{sen(2x)}{x}}\\\displaystyle\mathsf{\lim_{x \to 0}\dfrac{sen(2x)}{x}.\dfrac{2}{2}}\\\displaystyle\mathsf{2.\lim_{x \to 0}\dfrac{sen(2x)}{2x}}$

realizando uma substituição adequada temos:

$\mathsf{u=2x}\\\mathsf{u\to 0~quando~x\to 0}$

$\displaystyle\mathsf{2.\lim_{x \to 0}\dfrac{sen(2x)}{2x}}=\displaystyle\mathsf{2.\lim_{u \to 0}\dfrac{sen(u)}{u}=2.1=2}$

portanto

$\displaystyle\mathsf{f'(0)=\lim_{x \to 0}\dfrac{x+sen(2x)}{x}=1+2=3}$