Question

Proof that if f(x)=x^n, thenf'(x)=nx^n-1​

Answer 1

The proof that if
`(f(x) = x^n)`, then
`(f'(x) = nx^(n-1))` is done using the definition of the derivative and the binomial theorem.

Here's the proof:

Definition of the derivative: The derivative of a function (f(x)) is given by
`(f'(x) = \lim_(h \to 0) (f(x + h) - f(x))/(h)).`

For the function
`(f(x) = x^n)`, this becomes:

`[f'(x) = \lim_(h \to 0) ((x + h)^n - x^n)/(h)]`

Binomial theorem expansion: We can expand
`((x + h)^n)` using the binomial theorem:

`[(x + h)^n = x^n + nx^(n-1)h + (n(n-1))/(2)x^(n-2)h^2 + \ldots]`

Substituting this expansion into the expression for
`(f'(x))`, we get:

`[f'(x) = \lim_(h \to 0) (x^n + nx^(n-1)h + (n(n-1))/(2)x^(n-2)h^2 + \ldots - x^n)/(h)]`

Simplifying, we find that all terms with (h) in the numerator cancel out, except for the term
`(nx^(n-1))`. Dividing by (h) and taking the limit as (h \to 0), we get:

`[f'(x) = nx^(n-1)]`

This completes the proof that if
`(f(x) = x^n), then (f'(x) = nx^(n-1))`. The method used in this proof is a combination of applying the definition of the derivative and using the binomial theorem to expand the expression for the derivative.