Question

How do you find f'(x) using the limit definition given `f(x) = (3 + x)/(1 - 3x)`?

Answer 1

Substitute `f(x+h)` and `f(x)` into the definition.
Make a common denominator.
Combine like terms
Cancel the h from the denominator by division.
Let `hto0`

Explanation:

`f'(x) = lim_(hrarr0) (f(x+h)-f(x))/h" [1]"`

Given: `f(x) = (x+3)/(1-3x)" [2]"`

`f(x+h) = (x+h+3)/(1-3x-3h)" [3]"`

Substitute equations [2] and [3] into equation [1]:

`f'(x) = lim_(hrarr0) ((x+h+3)/(1-3x-3h)-(x+3)/(1-3x))/h`

We need to make a common denominator within the numerator:

`f'(x) = lim_(hrarr0) ((x+h+3)/(1-3x-3h)(1-3x)/(1-3x)-(x+3)/(1-3x)(1-3x-3h)/(1-3x-3h))/h`

Multiply the numerators within the numerator:

`f'(x) = lim_(hrarr0) ((x+h+3-3x^2-3xh-9x)/((1-3x-3h)(1-3x))-(x+3-3x^2-9x-3xh-9h)/((1-3x)(1-3x-3h)))/h`

Combine over 1 denominator:

`f'(x) = lim_(hrarr0) (((x+h+3-3x^2-3xh-9x)-(x+3-3x^2-9x-3xh-9h))/((1-3x)(1-3x-3h)))/h`

Everything sums to 0 except the h terms:

`f'(x) = lim_(hrarr0) ((10h)/((1-3x-3h)(1-3x)))/h`

Please observe the h cancels by division:

`f'(x) = lim_(hrarr0) ((10cancel(h))/((1-3x-3h)(1-3x)))/cancel(h)`

`f'(x) = lim_(hrarr0) 10/((1-3x-3h)(1-3x))`

It is safe to let `hrarr0`:

`f'(x) = 10/((1-3x)(1-3x)) = 10/(1-3x)^2`