How to find the quadratic approximation of a quotient? How would I find the quadratic approximation of some quotient like tan(x)?  
Can I rewrite it as $\tan(x) = \frac{\sin(x)}{\cos(x)}$, and then find the approximation for sine and cosine separately?
 A: You can do it; this is series composition.
Suppose that you want the Taylor series of $$y=\frac{f(x)}{g(x)}$$ built around $x=0$ up to second order. You have
$$f(x)=f(0)+x f'(0)+\frac{1}{2} x^2 f''(0)+O\left(x^3\right)$$
$$g(x)=g(0)+x g'(0)+\frac{1}{2} x^2 g''(0)+O\left(x^3\right)$$
$$y=\frac{f(0)+x f'(0)+\frac{1}{2} x^2 f''(0)+O\left(x^3\right) }{g(0)+x g'(0)+\frac{1}{2} x^2 g''(0)+O\left(x^3\right) }$$ Now, use the long division to get 
$$y=\frac{f(0)}{g(0)}+\frac{ \left(g(0) f'(0)-f(0) g'(0)\right)}{g(0)^2}x+\frac{
   \left(g(0)^2 f''(0)-2 g(0) f'(0) g'(0)-f(0) g(0) g''(0)+2 f(0) g'(0)^2\right)}{2 g(0)^3}x^2+O\left(x^3\right)$$
A: I’m not entirely sure if this is what you’re asking, but I suppose you could use the first few terms of the Taylor series
$$\tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} +\cdots $$
for $|x|<\pi/2$. Does that help?
A: $$\begin{align}
f(x) &= \tan x \\
f’(x) &= \sec^2x \\
f’’(x) &= 2\sec(x)\tan(x) \\
\end{align}$$
You should just memorize $f’(x)$, and $f’’(x)$ comes from chain rule.
Assuming we expand the approximation around $x=a$, we’ll use
$$f(x)\approx f(0)+f’(0)\,(x-a)+\frac{f’’(0)\,(x-a)^2}{2}$$
Use
$$\begin{align}
\tan0&=0\\
\sec^20&=1\\
2\sec(0)\tan(0)&=0\\
\end{align}\\$$
Giving you the rather lousy $f(x)\approx x$, hence my recommendation in my other answer.
