Question

calcular f'(x) pela definição, sendo, y= X² + X , X=1/2

Answer 1

Resposta:

Explicação passo-a-passo:

$f'(x)= \lim_{x\to a}\dfrac{f(x)-f(a)}{x-a}\\\\\\f'(\frac{1}{2})= \lim_{x\to \frac{1}{2}}\dfrac{x^{2}+x-((\frac{1}{2})^{2}+\frac{1}{2})}{x-\frac{1}{2}}\\\\\\f'(\frac{1}{2})= \lim_{x\to \frac{1}{2}}\dfrac{x^{2}+x-(\frac{1}{4}+\frac{1}{2})}{x-\frac{1}{2}}\\\\\\f'(\frac{1}{2})= \lim_{x\to \frac{1}{2}}\dfrac{x^{2}+x-(\frac{1+2}{4})}{x-\frac{1}{2}}\\\\\\f'(\frac{1}{2})= \lim_{x\to \frac{1}{2}}\dfrac{x^{2}+x-(\frac{3}{4})}{x-\frac{1}{2}}\\\\obs\\\\x^{2}+x-(\frac{3}{4})=(x-\frac{1}{2})(x+\frac{3}{2})$

$f'(\frac{1}{2})= \lim_{x\to \frac{1}{2}}\dfrac{(x-\frac{1}{2})(x+\frac{3}{2})}{(x-\frac{1}{2})}\\\\f'(\frac{1}{2})= \lim_{x\to \frac{1}{2}}(x+\frac{3}{2})\\\\f'(\frac{1}{2})=\frac{1}{2}+\frac{3}{2}=\frac{4}{2}=2$