Question

How do you find the derivative of `f(x) = x^4` using the limit definition?

Answer 1

See below for using either of 2 definitions.

Explanation:

Using `lim_(hrarr0) (f(x+h)-f(x))/h`

You'll need to expand `(x+h)^4`. Your choices are to multiply

`(x+h)(x+h)(x+h)(x+h)`,

or to use the binomial expansion (with Pascal's triangle if that's how you learned it.)

Either way, you will get

`(x+h)^4 = x^4 + 4x^3h+6x^2h^2+4xh^3+h^4`.

So we have:

`lim_(hrarr0) (f(x+h)-f(x))/h = lim_(hrarr0) ((x+h)^4-x^4)/h`

`= lim_(hrarr0) (x^4 + 4x^3h+6x^2h^2+4xh^3+h^4 - x^4)/h`

`= lim_(hrarr0) (4x^3h+6x^2h^2+4xh^3+h^4)/h`

`= lim_(hrarr0) (cancelh(4x^3+6x^2h^2+4xh^2+h^3))/cancelh`

`= 4x^3`

Using`lim_(trarrx) (f(t)-f(x))/(t-x)`

You'll need to factor `t^4-x^4`. Since, `t=x` make this polynomial eauql to `0`, it must have `t-x` as a linear factor. Using trial and error or polynomial division, we get:

`t^4-x^4 = (t-x)(t^3+t^2x+tx^2+x^3)`.

We have:

`lim_(trarrx) (f(t)-f(x))/(t-x) = lim_(trarrx) (t^4- x^4)/(t-x)`

`= lim_(trarrx) (cancel((t-x))(t^3+t^2x+tx^2+x^3))/cancel((t-x))`

`= lim_(trarrx) (t^3+color(red)(t^2x)+color(green)(tx^2)+color(crimson)(x^3))`

`= x^3+color(red)(x^2x)+color(green)(x x^2)+color(crimson)(x^3)`

`= x^3+color(red)(x^3)+color(green)(x^3)+color(crimson)(x^3)`

`= 4x^3`