Question

Calcule a derivada da função f(x)=x²+a, no ponto x0=1, por meio do cálculo do limite f'(x0) = lim h->0 f(x0+h) - f(x0)/h

Answer 1

lembrando o produto notavel

$\boxed{(a+b)^2=a^2+2ab+b^2}$

:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::

$\boxed{\boxed{f'(x_0)= \lim_{h \to 0} \frac{f(x_0+h)-f(x_0)}{h} }}}$


aplicando isso no problema com x0=1
$f'(1)= \lim_{h \to 0} \frac{f(1+h)-f(1)}{h}$

temos:
$\bmatrix f(x)=x^2+a\\\\ f(1)=1^2+a= 1+a\\\\\\f(1+h)=(1+h)^2+a=(1^2+2*1*h+h^2)+a = 1+2h+h^2+a\end$

colocando isso no limite:

$f'(1)= \lim_{h \to 0} \frac{f(1+h)-f(1)}{h}\\\\ f'(1)= \lim_{h \to 0} \frac{1+2h+h^2+a-(1+a)}{h}\\\\ f'(1)= \lim_{h \to 0} \frac{\not 1+2h+h^2+\not a-\not 1-\not a}{h}\\\\ f'(1)= \lim_{h \to 0} \frac{2h+h^2}{h} \\\\ f'(1)= \lim_{h \to 0} \frac{h(2+h)}{h} \\\\ f'(1)= \lim_{h \to 0} \frac{\not h(2+h)}{ \not h} \\\\ f'(1)= \lim_{h \to 0} (2+h)\\\\f'(1)=2+0=2$

Answer 2

Caso esteja pelo app, e tenha problemas para visualizar esta resposta, experimente abrir pelo navegador https://brainly.com.br/tarefa/12238780

                                                                                                                               

Derivada no ponto

$\Large\boxed{\boxed{\boxed{\boxed{\displaystyle\sf f'(a)=\lim_{x \to a}\dfrac{f(x)-f(a)}{x-a}}}}}}$

$\sf f(x)=x^2+a\\\sf f(1)=1^2+a=1+a\\\displaystyle\sf f'(1)=\lim_{x \to1}\dfrac{x^2+a-(1+a)}{x-1}\\\displaystyle\sf f'(1)=\lim_{x \to 1}\dfrac{x^2+\diagup\!\!\!a-1-\diagup\!\!\!a}{x-1}\\\displaystyle\sf f'(1)=\lim_{x \to 1}\dfrac{x^2-1}{x-1}$

$\tt x^2-1=(x-1)\cdot(x+1)\\\displaystyle\sf f'(1)=\lim_{x \to 1}\dfrac{\diagup\!\!\!\!(x-\diagup\!\!\!\!\!1)\cdot(x+1)}{\diagup\!\!\!\!\!(x-\diagup\!\!\!\!\!1)}\\\sf\displaystyle\sf f'(1)= \lim_{x \to 1}x+1=1+1\\\huge\boxed{\boxed{\boxed{\boxed{\displaystyle\sf f'(1)=2}}}}\checkmark$