How to use the first principle in finding derivative of f(x)=square root (1-x)?

We'll recall the first
principle:

lim [f(x+h) - f(x)]/h, for
h->0

Comparing, we'll
get:

lim {sqrt [1 - (x+h)] -
sqrt(1-x)}/h

We'll remove the brackets at
radicand:

lim [sqrt (1 - x - h) - sqrt(1 -
x)]/h

We'll multiply both, numerator and denominator, by
the conjugate of numerator:

lim [sqrt (1 - x - h) - sqrt(1
- x)]*[sqrt (1 - x - h) + sqrt(1 - x)]/h*[sqrt (1 - x - h) + sqrt(1 -
x)]

The product at numerator returns the difference of
squares:

lim [(1  -x - h) - (1  -x)]/h*[sqrt (1 - x - h) +
sqrt(1 - x)]

We'll eliminate like terms form
numerator:

lim -h/h*[sqrt (1 - x - h) + sqrt(1 -
x)]

We'll simplify and we'll
get:

lim -1/[sqrt (1 - x - h) + sqrt(1 -
x)]

We'll replace h by 0:

lim
-1/[sqrt (1 - x - h) + sqrt(1 - x)] = -1/[sqrt (1 - x) + sqrt(1 -
x)]

We'll combine like terms from
denominator:

lim -1/[sqrt (1 - x - h) + sqrt(1 - x)] =
-1/2sqrt (1 - x)

The first derivative of the
given function, using the first principle, is f'(x) = -1/2sqrt (1 -
x).