Solución

ddx(x3)=limh0(x+h)3x3h=limh0x3+3x2h+3xh2+h3x3hEl primer teˊrmino de    (x+h)3   es   x3=limh03x2h+3xh2+h3hCancelando   x3=limh0h(3x2+3xh+h2)hSacando factor comuˊn=limh0(3x2+3xh+h2)Simplificando=3x2Calculando el lıˊmite\begin{aligned} \frac{d}{dx} (x^3) &= \mathop {\lim }\limits_{h \to 0} \frac{(x + h)^3 − x^3}{h } \\ &= \mathop {\lim }\limits_{h \to 0} \frac{x^3 + 3x^2h + 3xh^2 + h^3 − x^3}{h } &\text{El primer término de }\,\,\,(x+h)^3\,\,\,\text{es}\,\,\,x^3\\ &= \mathop {\lim }\limits_{h \to 0} \frac{3x^2h + 3xh^2 + h^3}{h} &\text{Cancelando}\,\,\,x^3\\ &= \mathop {\lim }\limits_{h \to 0} \frac{h (3x^2 + 3xh + h^2)}{ h} &\text{Sacando factor común}\\ &= \mathop {\lim }\limits_{h \to 0} (3x^2 + 3xh + h^2) &\text{Simplificando}\\ &= 3x^2 &\text{Calculando el límite}\\ \end{aligned}