Question

Using the limit definition, how do you find the derivative of `f(x) =sqrt (x+1)`?

Answer 1

To find the derivative, apply the definition:

`f'(x) = lim_(h->0) (f(x+h) - f(x))/h`

Explanation:

With `f(x) = sqrt(x+1)`, we apply the definition:

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

To handle the terms in the numerator, it looks like we will need to multiply by the conjugate:

`f'(x) =`
`lim_(h->0) (sqrt(x+h+1) - sqrt(x+1))/h * (sqrt(x+h+1) + sqrt(x+1))/(sqrt(x+h+1) + sqrt(x+1))`

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

`f'(x) = lim_(h->0) cancel(h)/(cancel(h)(sqrt(x+h+1) + sqrt(x+1)))`

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

At this point, we can directly apply the limit to arrive at an answer:

`f'(x) = 1/(sqrt(x + 0 + 1) + sqrt(x + 1))`

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

`f'(x) = 1/(2sqrt(x+1))`

If you are familiar with the chain rule for derivatives, we can use it to test our result:

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