Question

If `f(x)=ax^n`, then why is `f'(x)=nax^(n-1)`?

Answer 1

By definition of the derivative:

`f'(x) = lim_(h->0) (f(x+h)-f(x))/h`

If `f(x) = ax^n` then we have:

`f'(x) = lim_(h->0) (a(x+h)^n-ax^n)/h = a lim_(h->0) 1/h((x+h)^n-x^n)`

Now use the binomial formula:

`f'(x) = a lim_(h->0) 1/h (sum_(j=0)^n ((n),(j)) x^(n-j) h^j -x^n)`

`f'(x) = a lim_(h->0) 1/h (cancel(x^n)+nx^(n-1)h + (n(n-1))/2 x^(n-2)h^2+...nxh^(n-1)+h^n-cancel(x^n))`

The two terms in `x^n` cancel each other and all the other have a factor that is a power of `h`:

`f'(x) = a lim_(h->0) 1/cancelh (cancelh (nx^(n-1) + (n(n-1))/2 x^(n-2)h+...nxh^(n-2)+h^(n-1))`

`f'(x) = a lim_(h->0) (nx^(n-1) + (n(n-1))/2 x^(n-2)h+...nxh^(n-2)+h^(n-1))`

Now for `h->0` all the powers of `h` go to zero, so:

`f'(x) = anx^(n-1)`