How to differentiate y=x^n using the first principle. In this question, I cannot use the rule of differentiation. I have to do the old fashioned way. How do I do it?

Question

asked Dec 22, 2020 167k views

1 Answer

Imposeren · Answer 1 · 2020-12-27T13:01:05+0000

By first principles, the derivative is

$\displaystyle\lim_(h\to0)\frac{(x+h)^n-x^n}h$

Use the binomial theorem to expand the numerator:

$(x+h)^n=\displaystyle\sum_(i=0)^n\binom nix^(n-i)h^i=\binom n0x^n+\binom n1x^(n-1)h+\cdots+\binom nnh^n$

$(x+h)^n=x^n+nx^(n-1)h+\frac{n(n-1)}2x^(n-2)h^2+\cdots+nxh^(n-1)+h^n$

where

$\dbinom nk=(n!)/(k!(n-k)!)$

The first term is eliminated, and the limit is

$\displaystyle\lim_(h\to0)\frac{nx^(n-1)h+\frac{n(n-1)}2x^(n-2)h^2+\cdots+nxh^(n-1)+h^n}h$

A power of
in every term of the numerator cancels with
in the denominator:

$\displaystyle\lim_(h\to0)\left(nx^(n-1)+\frac{n(n-1)}2x^(n-2)h+\cdots+nxh^(n-2)+h^(n-1)\right)$

Finally, each term containing
approaches 0 as
$h\to0$ , and the derivative is

$y=x^n\implies y'=nx^(n-1)$