Question

Derivative Question help!?

How do you find the derivative of f(x) = `sqrt(x-7)` using definition of the derivative?

Answer 1

Given: `f(x) = sqrt(x-7)`, then `f(x+h) = sqrt(x + h-7)`

Using the definition, `f'(x) = lim_(h to 0) (f(x+h)-f(x))/h`:

`f'(x) = lim_(h to 0) (sqrt(x + h-7)-sqrt(x-7))/h`

We need to eliminate the radicals in the numerator and we know that `(a-b)(a+b) = a^2-b^2`, therefore, we shall multiply by 1 in the form `(sqrt(x + h-7)+sqrt(x-7))/(sqrt(x + h-7)+sqrt(x-7))`:

`f'(x) = lim_(h to 0) (sqrt(x + h-7)-sqrt(x-7))/h(sqrt(x + h-7)+sqrt(x-7))/(sqrt(x + h-7)+sqrt(x-7))`

We know that this will eliminate the radicals in the numerator by squaring them and the denominator is just multiplication:

`f'(x) = lim_(h to 0) (x + h-7-(x-7))/(h(sqrt(x + h-7)+sqrt(x-7)))`

All of the terms in the numerator sum to 0 except for h:

`f'(x) = lim_(h to 0) h/(h(sqrt(x + h-7)+sqrt(x-7)))`

`h/h` becomes 1:

`f'(x) = lim_(h to 0) 1/(sqrt(x + h-7)+sqrt(x-7))`

Now, we can allow h to become 0:

`f'(x) = 1/(sqrt(x-7)+sqrt(x-7))`

Simplify:

`f'(x) = 1/(2sqrt(x-7))`