Question

seja f(x) = x² + 5 x -1, derive usando a definição

Answer 1

$f(x)=x^{2}+5x-1$


A derivada de $f$ por definição é o resultado do cálculo deste limite:


(nos pontos onde o limite abaixo existir)

$f'(x)=\underset{h\to 0}{\mathrm{\ell im}}~\dfrac{f(x+h)-f(x)}{h}\\ \\ \\ =\underset{h\to 0}{\mathrm{\ell im}}~\dfrac{\left[(x+h)^{2}+5(x+h)-1 \right ]-(x^{2}+5x-1)}{h}\\ \\ \\ =\underset{h\to 0}{\mathrm{\ell im}}~\dfrac{\left[(x^{2}+2xh+h^{2})+5(x+h)-1 \right ]-x^{2}-5x+1}{h}\\ \\ \\ =\underset{h\to 0}{\mathrm{\ell im}}~\dfrac{\diagup\!\!\!\!\!x^{2}+2xh+h^{2}+\diagup\!\!\!\diagup\!\!\!\!\!\!\! 5x+5h-\diagdown\!\!\!\! 1-\diagup\!\!\!\!\!x^{2}-\diagup\!\!\!\diagup\!\!\!\!\!\!\!5x+\diagdown\!\!\!\! 1}{h}$

$=\underset{h\to 0}{\mathrm{\ell im}}~\dfrac{2xh+h^{2}+5h}{h}\\ \\ \\ =\underset{h\to 0}{\mathrm{\ell im}}~\dfrac{\diagup\!\!\!\!h\cdot (2x+h+5)}{\diagup\!\!\!\! h}\\ \\ \\ =\underset{h\to 0}{\mathrm{\ell im}}~(2x+h+5)\\ \\ \\ =2x+0+5\\ \\ =2x+5\\ \\ \\ \therefore~~\boxed{\begin{array}{l} f'(x)=2x+5 \end{array}}$