Question

Simplifique $\frac{f(x+h) - f(x)}{h}$

letra a - 1/x
letra b - 1/x²

Answer 1

$\dfrac{f(x+h)-f(x)}{h}~~~~~~\text{com }h\ne 0$

________

a) $f(x)=\dfrac{1}{x}$

$\dfrac{f(x+h)-f(x)}{h}\\\\\\ =\dfrac{\frac{1}{x+h}-\frac{1}{x}}{h}\\\\\\ =\left(\dfrac{1}{x+h}-\dfrac{1}{x} \right )\cdot \dfrac{1}{h}\\\\\\ =\left(\dfrac{x}{(x+h)\cdot x}-\dfrac{x+h}{(x+h)\cdot x} \right )\cdot \dfrac{1}{h}\\\\\\ =\dfrac{x-(x+h)}{(x+h)\cdot x}\cdot \dfrac{1}{h}$

$=\dfrac{\diagup\!\!\!\! x-\diagup\!\!\!\! x-h}{(x+h)\cdot x\cdot h}\\\\\\ =\dfrac{(-1)\cdot \diagup\!\!\!\! h}{(x+h)\cdot x\cdot \diagup\!\!\!\! h}\\\\\\ =\boxed{\begin{array}{c}-\,\dfrac{1}{(x+h)\cdot x} \end{array}}$

________

b) $f(x)=\dfrac{1}{x^2}$

$\dfrac{f(x+h)-f(x)}{h}\\\\\\ =\dfrac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}\\\\\\ =\left(\dfrac{1}{(x+h)^2}-\dfrac{1}{x^2} \right )\cdot \dfrac{1}{h}\\\\\\ =\left(\dfrac{x^2}{(x+h)^2\cdot x^2}-\dfrac{(x+h)^2}{(x+h)^2\cdot x^2} \right )\cdot \dfrac{1}{h}\\\\\\ =\dfrac{x^2-(x+h)^2}{(x+h)^2\cdot x^2}\cdot \dfrac{1}{h}$

$=\dfrac{x^2-(x^2+2xh+h^2)}{(x+h)^2\cdot x^2\cdot h}\\\\\\ =\dfrac{\diagup\!\!\!\!\! x^2-\diagup\!\!\!\!\! x^2-2xh-h^2}{(x+h)^2\cdot x^2\cdot h}\\\\\\ =\dfrac{\diagup\!\!\!\!\! h\cdot (-2x-h)}{(x+h)^2\cdot x^2\cdot \diagup\!\!\!\!\! h}\\\\\\ =\dfrac{-2x-h}{(x+h)^2\cdot x^2}\\\\\\ =\boxed{\begin{array}{c}-\,\dfrac{2x+h}{(x+h)^2\cdot x^2} \end{array}}$


Bons estudos! :-)