How do you use the definition of derivative in terms of limits, prove that the derivative of xn=nxn−1?

Question

izvozna39g0 · Accepted Answer

Definition of Derivativef′(x)=limh→0f(x+h)−f(x)hFor f(x)=xnlimh→0(x+h)n−xnhlimh→0xn+nh1xn−1+h2(…)−xnhlimh→0nxn−1+h(…)=nxn−1

tempur8x43 · Accepted Answer

Using the alternative definition of derivative:

f^{'} (a) = lim_{x \to a} \frac{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} + \dots + x a^{n - 2} + a^{n - 1})

So:

f^{'} (a) = lim_{x \to a} \frac{f (x) - f (a)}{x - a}

= lim_{x \to a} \frac{x^{n} - a^{n}}{x - a}

= lim_{x \to a} \frac{(x - a) (x^{n - 1} + x^{n - 2} a + x^{n - 3} a^{2} + \dots + x a^{n - 2} + a^{n - 1})}{x - a}

= lim_{x \to a} (x^{n - 1} + x^{n - 2} a + x^{n - 3} a^{2} + \dots + x a^{n - 2} + a^{n - 1})

= a^{n - 1} + a^{n - 2} + a^{n - 3} a^{2} + \dots + a^{n - 2} + a^{n - 1}

There are n factor, each of which has limit

a^{n - 1}

, so the limit is

f^{'} (a) = n a^{n - 1}

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

f^{'} (x) = n x^{n - 1}

How do you use the definition of derivative