Question

Using the limit definition, how do you find the derivative of `f ( x) = x^4`?

Answer 1

`f' (x)=4x^3`

Explanation:

Given `f (x)=x^4`

Let `y=x^4`

replace `y` with `y+Delta y` and `x` with `x+Delta x`

`y=x^4`

`y+Delta y=(x+Delta x)^4`

`y+Delta y=x^4+4*x^3(Delta x)+6*x^2(Delta x)^2+4x(Delta x)^3+(Delta x)^4`

Subtract `y` from both sides

`y+Delta ycolor(red)(-y)=x^4+4*x^3(Delta x)+6*x^2(Delta x)^2+4x(Delta x)^3+(Delta x)^4color(red)(-x^4)`

`Delta y=4*x^3(Delta x)+6*x^2(Delta x)^2+4x(Delta x)^3+(Delta x)^4`

Divide both sides by `Delta x`

`(Delta y)/(Delta x)=(4*x^3(Delta x)+6*x^2(Delta x)^2+4x(Delta x)^3+(Delta x)^4)/(Delta x)`

`(Delta y)/(Delta x)=4*x^3+6*x^2(Delta x)+4x(Delta x)^2+(Delta x)^3`

Take now the `color(blue)("LIMIT")` of `(Delta y)/(Delta x)` as `Delta x rarr 0`

`color(blue)(dy/dx=lim_(Deltax rarr 0)(Delta y)/(Delta x)=color(blue)(lim_(Delta xrarr 0)(4*x^3+6*x^2(Delta x)+4x(Delta x)^2+(Delta x)^3)=4x^3)`

God bless....I hope the explanation is useful.

Answer 2

Here is an alternative using one form of the limit definition.

Explanation:

I am using the limit definition in the form

`f'(a) = lim_(xrarra)(f(x)-f(a))/(x-a)`

For `f(x)=x^4`, we get

`f'(a) = lim_(xrarra)(x^4-a^4)/(x-a)` `" "`

(This has form `0/0`, so `x-a` is a factor of the numerator)

`= lim_(xrarra)((x-a)(x^3+x^2a+xa^2+a^3))/(x-a)` `" "` (form `0/0`)

`= lim_(xrarra)(x^3+x^2a+xa^2+a^3)`

`= (a)^3+(a)^2a+(a)a^2+a^3`

`= 4a^3`

Since `f'(a) = 4a^3`, we get

`f'(x) = 4x^3`