Expansion for gravitational time dilation

Question

In the section on gravitational time dilation in Prof. David Tong's lecture notes on general relativity, he performs the following expansion:

$$t\sqrt{1-\frac{2GM}{r_{A}c^{2}}+\frac{2GM{\Delta}r}{r^{2}_{A}c^{2}}}\\{\approx}t\sqrt{1-\frac{2GM}{r_{A}c^{2}}}\left(1+\frac{GM{\Delta}r}{r^{2}_{A}c^2}\right)$$

I was hoping someone could fill in the steps between the lines here, as I'm having some trouble.

Answer 1

I called the function f(x), so $x$ instead of $\Delta r$ and all divided by $t$.

$f(x) = \sqrt{1-\frac{2GM}{r_Ac^2}+\frac{2GM x}{r_A^2c^2}}$

$f'(x) = \frac{1}{2} \frac{1}{\sqrt{1-\frac{2GM}{r_Ac^2}+\frac{2GM x}{r_A^2c^2}}} \frac{2GM }{r_A^2c^2}$

With normal Taylor expansion around $x_0 = 0 $ to first order:

$f(x) \approx \sqrt{1-\frac{2GM}{r_Ac^2}} + \frac{1}{ \sqrt{1-\frac{2GM}{r_Ac^2}}}\frac{GM x}{r_A^2c^2}$

$ = \sqrt{1-\frac{2GM}{r_Ac^2}} (1+\frac{1}{1-\frac{2GM}{r_Ac^2}} \frac{GM x}{r_A^2c^2})$

If you say $\frac{1}{1-\frac{2GM}{r_Ac^2}} \approx 1$ you recover your result. But as for now, doing so does not make sense to me.