Question

Using the limit definition, how do you find the derivative of `f(x) = 3x^2 + 8x + 4`?

Answer 1

`f'(x)=6x+8`; see below for explanation

Explanation:

The limit definition of the derivative states that for a function `f(x)` its derivative equals

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

So, when `f(x)=3x^2+8x+4`, we see that `f(x+h)=3(x+h)^2+8(x+h)+4`.

Applying the limit definition, we obtain

`f'(x)=lim_(hrarr0)(3(x+h)^2+8(x+h)+4-(3x^2+8x+4))/h`

Find `(x+h)^2`. Distribute the negative into `-(3x^2+4x+8)`.

`f'(x)=lim_(hrarr0)(3(x^2+2hx+h^2)+8(x+h)+4-3x^2-8x-4)/h`

Distribute the `3` and the `8`.

`f'(x)=lim_(hrarr0)(3x^2+6hx+3h^2+8x+8h+4-3x^2-8x-4)/h`

Cancel all like terms.

`f'(x)=lim_(hrarr0)(color(red)(cancel(color(black)(3x^2)))+6hx+3h^2color(blue)(cancel(color(black)(+8x)))+8hcolor(green)(cancel(color(black)(+4)))color(red)(cancel(color(black)(-3x^2)))color(blue)(cancel(color(black)(-8x)))color(green)(cancel(color(black)(-4))))/h`

`f'(x)=lim_(hrarr0)(6hx+3h^2+8h)/h`

Divide `h` from each term.

`f'(x)=lim_(hrarr0)6x+3h+8`

To evaluate the limit, plug in `0` for `h`.

`f'(x)=6x+3(0)+8`

`f'(x)=6x+8`