Question

Como calcular a derivada pela definição, de g(x)$\frac{3-x}{2x+1}$

Answer 1

$g(x)=\dfrac{3-x}{2x+1}$

$\bullet\;\;$ Calculemos primeiro a razão incremental:

Fazendo $\Delta x=h,$ temos que

$\dfrac{\Delta g}{\Delta x}=\dfrac{g(x+\Delta x)-g(x)}{\Delta x}\\ \\ \\ \dfrac{\Delta g}{h}=\dfrac{g(x+h)-g(x)}{h}\\ \\ \\ \dfrac{\Delta g}{h}=\dfrac{\frac{3-(x+h)}{2(x+h)+1}-\frac{3-x}{2x+1}}{h}\\ \\ \\ \dfrac{\Delta g}{h}=\dfrac{1}{h}\cdot \left[\dfrac{3-(x+h)}{2(x+h)+1}-\dfrac{3-x}{2x+1} \right ]\\ \\ \\ \dfrac{\Delta g}{h}=\dfrac{1}{h}\cdot \dfrac{[3-(x+h)]\cdot (2x+1)-(3-x)\cdot [2(x+h)+1]}{[2(x+h)+1]\cdot (2x+1)}\\ \\ \\ \dfrac{\Delta g}{h}=\dfrac{1}{h}\cdot \dfrac{(3-x-h)\cdot (2x+1)-(3-x)\cdot (2x+2h+1)}{(2x+2h+1)\cdot (2x+1)}\\ \\ \\ \dfrac{\Delta g}{h}=\dfrac{1}{h}\cdot \dfrac{[(3-x)-h]\cdot (2x+1)-(3-x)\cdot [(2x+1)+2h]}{(2x+2h+1)\cdot (2x+1)}$


Aplicando a distributiva no numerador, temos

$\dfrac{\Delta g}{h}=\dfrac{1}{h}\cdot \dfrac{(3-x)\cdot (2x+1)-h\cdot (2x+1)-(3-x)\cdot (2x+1)-(3-x)\cdot 2h}{(2x+2h+1)\cdot (2x+1)}\\ \\ \\ \dfrac{\Delta g}{h}=\dfrac{1}{h}\cdot \dfrac{-h\cdot (2x+1)-(3-x)\cdot 2h}{(2x+2h+1)\cdot (2x+1)}$


Colocando $h$ em evidência no numerador, temos

$\dfrac{\Delta g}{h}=\dfrac{1}{\diagup\!\!\!\! h}\cdot \dfrac{\diagup\!\!\!\! h\cdot [-(2x+1)-(3-x)\cdot 2]}{(2x+2h+1)\cdot (2x+1)}\\ \\ \\ \dfrac{\Delta g}{h}=\dfrac{-2x-1-6+2x}{(2x+2h+1)\cdot (2x+1)}\\ \\ \\ \dfrac{\Delta g}{h}=\dfrac{-7}{(2x+2h+1)\cdot (2x+1)}$


$\bullet\;\;$ Por definição, a derivada é o limite da razão incremental, quando o incremento $h$ tende a zero:

$\dfrac{dg}{dx}=\underset{h\to 0}{\mathrm{\ell im}}\;\dfrac{\Delta g}{h}\\ \\ \\ =\underset{h\to 0}{\mathrm{\ell im}}\;\dfrac{-7}{(2x+2h+1)\cdot (2x+1)}\\ \\ \\ =\dfrac{-7}{(2x+0+1)\cdot (2x+1)}\\ \\ \\ =\dfrac{-7}{(2x+1)^{2}}$