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

Answer 1

According to the derivative's definition:

#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)#

Apply the binomial formula now:

#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)#
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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