Question

Calcule a derivada da função F(x) = x^n, usando a definição de derivada.

Answer 1

$\frac{df(x)}{dx}~=~^{\,lim}_{h\to0}~\frac{(x+h)^n~-~x^n}{h}\\\\\\Vamos~expandir~o~binomio~(x+h)^n~utilizando~o~Binomio~de~Newton\\\\\\(x+h)^n~=~\sum\limits_{k\,=\,0}^{n}~\binom{n}{k}\,.\,x^{n-k}\,.\,h^k\\\\\\(x+h)^n~=~x^n~+~n.x^{n-1}.h~+~\frac{n(n-1)}{2!}x^{n-2}h^2~+~\frac{n(n-1)(n-2)^}{3!}x^{n-3}h^3~+~...~+~h^n$

Substituindo na definição da derivada:

$\frac{df(x)}{dx}~=~^{\,lim}_{h\to0}~\frac{\left(x^n~+~n.x^{n-1}.h~+~\frac{n(n-1)}{2!}x^{n-2}h^2~+~\frac{n(n-1)(n-2)^}{3!}x^{n-3}h^3~+~...~+~h^n\right)-x^n}{h}\\\\\\O~termo~x^n~dentro~dos~parentesis~corta~com~o~x^n~de~fora\\\\\\\frac{df(x)}{dx}~=~^{\,lim}_{h\to0}~\frac{\left(n.x^{n-1}.h~+~\frac{n(n-1)}{2!}x^{n-2}h^2~+~\frac{n(n-1)(n-2)^}{3!}x^{n-3}h^3~+~...~+~h^n\right)}{h}\\\\\\Note~que~todos~termos~do~numerador~possuem~"h",~logo~podemos~\\cortar~com~o~"h"~do~denominador:$

$\frac{df(x)}{dx}~=~^{\,lim}_{h\to0}~\left(n.x^{n-1}~+~\frac{n(n-1)}{2!}x^{n-2}h~+~\frac{n(n-1)(n-2)^}{3!}x^{n-3}h^2~+~...~+~h^{n-1}\right)\\\\\\Agora~o~limite~pode~ser~calculado~sem~problemas\\\\\\\frac{df(x)}{dx}~=~\left(n.x^{n-1}~+~0~+~0~+~...~+~0\right)\\\\\\\boxed{\frac{df(x)}{dx}~=~n.x^{n-1}}$