Tangent line question

[yullin]

I am totally stuck on this problem which says:

. . . . .Find the linear approximation of the tangent line to:

. . . . . . . .f(x) = 1 / sqrt[1 - x]

. . . . .at x = 0

I have this:

. . . . .f(0) = 1/sqrt(1 - 0) = 1

. . . . .f(x) = 1/(1 - x)^(1/2)

so:

. . . . .f'(x) = (1-x)^(1/2)(0) - 1(1/2)(1-x)^(-1/2)(1) / (1-x)

I simplified it to:

. . . . .f'(x) = -1(1/2)(1 - x)^(-1/2)(1) / (1 - x)

I know that it sounds really stupid, but I don't know how to simplify it further.

 

[pka]

First the derivative is \(\displaystyle \L
f'(x) = \frac{1}{{2\sqrt {\left( {1 - x} \right)^3 } }}\).

The linear approximation of f at a is
\(\displaystyle \L
L(a) \approx f(a) + f'(a)(x - a)\)