Question

How do you use the definition of the derivative to find `f'(x)` if `f(x)=1/sqrtx`?

Answer 1

Start with `lim/(h→0){f(x+h)-f(x)}/h` multiply numerator and denominator by `f(x+h) + f(x)`

Explanation:

`lim/(h→0){(f(x+h))^2-(f(x))^2}/(h(f(x) + f(x+h)))`

We observe that `(f(x))^2 = 1/x` and `(f(x+h))^2 = 1/(x+h)` and we do the substitution:

`lim/(h→0){1/(x+h) - 1/x}/(h(1/sqrt(x) + 1/(sqrt(x+h)))`

Make a common denominator for `1/(x+h) - 1/x = (x)/((x(x +h))) - (x+h)/((x(x +h))) = -h/((x(x +h)))`

Substitute back into the limit:

`lim/(h→0){-h/((x(x +h)))}/(h(1/sqrt(x) + 1/(sqrt(x+h)))`

Please observe that `-h/h = -1`

`lim/(h→0){-1/((x(x +h)))}/((1/sqrt(x) + 1/(sqrt(x+h)))`

Now it is safe to let h go to zero:

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