Question

How do you find the derivative of `f(x) = 1/sqrt(2x-1)` by first principles?

Answer 1

derivative of function to a power `-1/2 2(2x-1)^(-1/2-1)`

Explanation:

`f(x) = (2x-1)`
You want the derivative of f to the power `-1/2`
Chain rule
`-1/2 2(2x-1)^(-1/2-1)`

The factor 2 comes from the derivative of f itself

Answer 2

Use limit definition of derivative to find:

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

Explanation:

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

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

`=lim_(h->0) (1/sqrt(2a+2h-1) - 1/sqrt(2a-1))/h`

`=lim_(h->0) (sqrt(2a-1)-sqrt(2a+2h-1))/(h sqrt(2a+2h-1) sqrt(2a-1))`

`=lim_(h->0) ((sqrt(2a-1)-sqrt(2a+2h-1))(sqrt(2a-1)+sqrt(2a+2h-1)))/(h sqrt(2a+2h-1) sqrt(2a-1) (sqrt(2a-1)+sqrt(2a+2h-1)))`

`=lim_(h->0) ((2a-1)-(2a+2h-1))/(h sqrt(2a+2h-1) sqrt(2a-1) (sqrt(2a-1)+sqrt(2a+2h-1)))`

`=lim_(h->0) (-2h)/(h sqrt(2a+2h-1) sqrt(2a-1) (sqrt(2a-1)+sqrt(2a+2h-1)))`

`=lim_(h->0) (-2)/(sqrt(2a+2h-1) sqrt(2a-1) (sqrt(2a-1)+sqrt(2a+2h-1)))`

`=(-2)/(sqrt(2a-1) sqrt(2a-1) (sqrt(2a-1)+sqrt(2a-1)))`

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

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

So:

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