Consider the function $f$:
$$f(v) = \left(1-\dfrac{v^2}{c^2}\right)^{\frac{-1}{2}}$$
Computing linear approximation to f as $v \approx 0$ - Direct differentiation
$$f'(v) = \dfrac{-1}{2} \cdot \left(1 - \dfrac{v^2}{c^2}\right)^{\frac{-3}{2}} \cdot \dfrac{-2v}{c^2}$$
Then, $f'(0) = 0$.
Then, $f(v) \approx f(0) + f'(0) \cdot v = 1$, as $v \approx 0$.
Computing linear approximation to f as $u \approx 0$ - Substitution
Let $u = \dfrac{-v^2}{c^2}$. Then, $f(u) = (1+u)^{\frac{-1}{2}}$.
Then, $f'(u) = \dfrac{-1}{2} \cdot (1+u)^{\frac{-3}{2}}$. Therefore, $f'(0) = \dfrac{-1}{2}$.
Then, $f(u) \approx f(0) + f'(0) \cdot u = 1 + \dfrac{-1}{2} \cdot u.$
Then, $f(v) \approx 1+\dfrac{1}{2}\left(\dfrac{v^2}{c^2}\right)$, as $v \approx 0$.
The resource I'm studying from, lists the second method as giving the correct linear approximation. Where am I going wrong then, with the first method?
The "linear" is in reference to $v^2$, not $v$, as the base variable. The Taylor expansion of $f$ around $0$ has only even powers of $v$, and you only calculated to the first power of $v$, so you only got a constant.
$f$ is the expression for the Lorentz factor $\gamma$ in special relativity.