Question

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

Answer 1

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

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)`