Question

Ayudaaa con estos ejercicioss Derivada por Incremento doy todosss mis puntoooosss porfavor es Urgenteee!

Answer 1

(1)
$f(x)=ax^2\\ \\ f(x+\Delta x)= a(x+\Delta x)^2\\ f(x+\Delta x)= a[x^2+2x\Delta x+(\Delta x)^2]\\ f(x+\Delta x)= ax^2+2ax\Delta x+a(\Delta x)^2\\ \\ \Delta y = f(x+\Delta x)-f(x)\\ \Delta y =ax^2+2ax\Delta x+a(\Delta x)^2-ax^2\\ \Delta y =2ax\Delta x+a(\Delta x)^2\\ \\ \displaystyle \frac{\Delta y}{\Delta x} =2ax+a\cdot\Delta x$

(2)
$\displaystyle g(x)=\frac{x^2}{4-x^2}\\ \\ g(x+\Delta x)=\frac{(x+\Delta x)^2}{4-(x+\Delta x)^2}\\ \\ \\ \Delta y = g(x+\Delta x)-g(x)\\ \\ \Delta y = \frac{(x+\Delta x)^2}{4-(x+\Delta x)^2}-\frac{x^2}{4-x^2}\\ \\ \Delta y = \frac{x^2+2x\Delta x+(\Delta x)^2}{4-(x^2+2x\Delta x+(\Delta x)^2)}-\frac{x^2}{4-x^2}\\ \\ \Delta y = \frac{(x^2+2x\Delta x+(\Delta x)^2)(4-x^2)-x^2[4-(x^2+2x\Delta x+(\Delta x)^2)]}{[4-(x^2+2x\Delta x+(\Delta x)^2)](4-x^2)}$

$\displaystyle \Delta y=\frac{4\Delta x(2x + \Delta x)}{(x^2 - 4)(x^2 + 2x\Delta x + (\Delta x)^2 - 4)}\\ \\ \\ \frac{\Delta y}{\Delta x}=\frac{8x + 4\Delta x}{(x^2 - 4)^2+(x^2 - 4)(2x + \Delta x)\Delta x}$