Question

How do you find f'(x) using the limit definition given `f(x)=x^(-1/2)`?

Answer 1

see below

Explanation:

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

Here we have

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

`= lim_(h to 0) 1/h * (( sqrt(x) - sqrt(x+h) )/(sqrt(x+h) sqrt(x)))`

A very common tactic when dealing with radicals is to multiply by the conjugate, and use the idea that `(a+b)(a-b) = a^2 - b^2`

Here that means:

`= lim_(h to 0) 1/h * (( sqrt(x) - sqrt(x+h) )/(sqrt(x+h) sqrt(x)))* (( sqrt(x) + sqrt(x+h) )/( sqrt(x) + sqrt(x+h)))`

`= lim_(h to 0) 1/h * (( (x) - (x+h) )/(x sqrt(x+h) + (x+h) sqrt(x)))`

doing the numerator first
`= lim_(h to 0) ( -1 )/(x sqrt(x+h) + (x+h) sqrt(x))`

then the denominator
`= lim_(h to 0) - ( 1 )/(x sqrt(x+h) + x sqrt(x) + h sqrt(x))`

`= - ( 1 )/(x sqrt(x) + x sqrt(x) + 0)`

`= - ( 1 )/(2x sqrt(x) )`

by some more manipulation, this then becomes the same as you would expect from a straight power rule

`= - ( 1 )/(2 sqrt(x^3) )`

`= - ( 1 )/(2 x^(3/2) )`