Question

How do you use the limit definition of the derivative to find the derivative of `f(x)=sqrt(x-4)`?

Answer 1

Please see the explanation for details on how you do the requested process.

Explanation:

The limit definition is:

`lim_(hto0){f(x+h) - f(x)}/h`

Given:

`f(x) = sqrt(x - 4)`

To obtain `f(x + h),`substitute `x + h` for `x`:

`f(x + h) = sqrt(x + h - 4)`

Substituting into the definition:

`lim_(hto0){sqrt(x + h - 4) - sqrt(x - 4)}/h`

Because we know that `(a - b)(a + b) = a^2 - b^2`, we choose to multiply by `{sqrt(x + h - 4) + sqrt(x - 4)}/{sqrt(x + h - 4) + sqrt(x - 4)}`:

`lim_(hto0){sqrt(x + h - 4) - sqrt(x - 4)}/h{sqrt(x + h - 4) + sqrt(x - 4)}/{sqrt(x + h - 4) + sqrt(x - 4)}`

This will square the square roots in the numerator and, eventually, leave nothing but h:

`lim_(hto0){(sqrt(x + h - 4))^2 - (sqrt(x - 4))^2}/(h{sqrt(x + h - 4) + sqrt(x - 4)})`

Squaring the square roots makes them disappear:

`lim_(hto0){x + h - 4 - (x - 4)}/(h{sqrt(x + h - 4) + sqrt(x - 4)})`

Distribute the - through the ()s in the numerator:

`lim_(hto0){x + h - 4 - x + 4}/(h{sqrt(x + h - 4) + sqrt(x - 4)})`

The numerator simplifies to become only h:

`lim_(hto0)h/(h{sqrt(x + h - 4) + sqrt(x - 4)})`

`h/h` becomes 1:

`lim_(hto0)1/{sqrt(x + h - 4) + sqrt(x - 4)}`

Now, it is ok to let `hto0`:

`1/{sqrt(x - 4) + sqrt(x - 4)}`

Combine the terms in the denominator:

`1/{2sqrt(x - 4)}`