Question

How do you find the derivative of `f(x)=sqrt(x+1)` using the limit process?

Answer 1

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

Find a substitution for h that cancels the numerator and allows the limit to go to zero.

Explanation:

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

Given: `f(x) = sqrt(x + 1)`, then `f(x + h) = sqrt(x + h + 1)`

For integer roots of any function, we can find a substitution for h:

`f(x + h) = sqrt(x + h + 1)`

Square both sides:

`(f(x + h))^2 = x + h + 1`

Regroup the right side:

`(f(x + h))^2 = h + (x + 1)`

Please notice that what is inside the ()s is `(f(x))^2`:

`(f(x + h))^2 = h + (f(x))^2`

`h = (f(x + h))^2 - (f(x))^2`

This factors:

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

When we make the substitution for h, the numerator will cancel the left factor:

`lim_(hto0)cancel(f(x + h) - f(x))/(cancel(f(x + h) - f(x))(f(x + h) + f(x)))`

Here is the limit with the canceled factors eliminated:

`lim_(hto0)1/(f(x + h) + f(x))`

We can let the limit to go zero:

`1/(f(x) + f(x))`

`1/(2f(x))`

`1/(2sqrt(x + 1)`

This works for `sqrt, root(3), root(4), ...`. As long as it is an integer root, a substitution for h can be found that cancels the numerator and allows limit to go to zero.