How do you use the definition of derivative in terms of limits, prove that the derivative of $x^n = nx^(n-1)$ ?

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Answer 1 · 2015-04-29T16:22:31

Definition of Derivative
$f'(x) = lim_(hrarr 0) (f(x+h)-f(x))/h$

For $f(x) = x^n$

$lim_(hrarr0) ((x+h)^n - x^n)/h$

$lim_(hrarr0) (x^n +nh^1x^(n-1) + h^2(...) -x^n)/h$

$lim_(harr0) nx^(n-1) + h(...)$

$= nx^(n-1)$

Answer 2 · 2015-04-29T19:02:49

Using the alternative definition of derivative:

$f'(a) = lim_(xrarra) (f(x)-f(a))/(x-a)$

For $f(x) = x^n$ with $n$ a positive integer:

Note that:
$x^n-a^n = (x-a)(x^(n-1)+x^(n-2)a+x^(n-3)a^2 + * * * +xa^(n-2)+a^(n-1))$

So:

$f'(a) = lim_(xrarra) (f(x)-f(a))/(x-a)$

$color(white)"sssss"$ $= lim_(xrarra) (x^n-a^n)/(x-a)$

$=lim_(xrarra) ((x-a)(x^(n-1)+x^(n-2)a+x^(n-3)a^2 + * * * +xa^(n-2)+a^(n-1)))/(x-a)$

$=lim_(xrarra) (x^(n-1)+x^(n-2)a+x^(n-3)a^2 + * * * +xa^(n-2)+a^(n-1))$

$=a^(n-1)+a^(n-2)a+a^(n-3)a^2 + * * * +aa^(n-2)+a^(n-1)$

There are $n$ factor, each of which has limit $a^(n-1)$ , so the limit is

$f'(a) = na^(n-1)$

Because this holds for arbitrary $a$ , it holds for all $x$ and

$f'(x) = nx^(n-1)$

How do you use the definition of derivative in terms of limits, prove that the derivative of x^n = nx^(n-1)x^n = nx^(n-1)?