To find the first four terms of the Maclaurin series for $f(x) = \tan(x)$, we need to compute the derivatives of $f(x) = \tan(x)$ at $x = 0$ and use the formula for the Maclaurin series:

$f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots$

Step-by-Step Process:

1. Calculate the function and its derivatives at $x = 0$:

   Let's find the first few derivatives of $f(x) = \tan(x)$.

   • $f(x) = \tan(x)$

   • First derivative: $f'(x) = \sec^2(x)$

   • Second derivative: $f''(x) = 2 \sec^2(x) \tan(x)$

   • Third derivative: $f'''(x) = 2 \sec^2(x) (1 + 2 \tan^2(x))$

   • Fourth derivative: $f^{(4)}(x) = 8 \sec^2(x) \tan(x) (1 + \tan^2(x))$

2. Evaluate these derivatives at $x = 0$:

   At $x = 0$, we know:

   • $\tan(0) = 0$
   • $\sec(0) = 1$

   Therefore:

   • $f(0) = \tan(0) = 0$

   • $f'(0) = \sec^2(0) = 1$

   • $f''(0) = 2 \sec^2(0) \tan(0) = 2 \cdot 1 \cdot 0 = 0$

   • $f'''(0) = 2 \sec^2(0) (1 + 2 \tan^2(0)) = 2 \cdot 1 \cdot (1 + 0) = 2$

   • $f^{(4)}(0) = 8 \sec^2(0) \tan(0) (1 + \tan^2(0)) = 8 \cdot 1 \cdot 0 = 0$

3. Construct the Maclaurin series using these values:

   Substitute the values of the derivatives into the Maclaurin series formula:

$f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4 + \cdots$

Substituting the values we found:

$\tan(x) \approx 0 + 1 \cdot x + \frac{0}{2!}x^2 + \frac{2}{3!}x^3 + \frac{0}{4!}x^4 + \cdots$

Simplify:

$\tan(x) \approx x + \frac{x^3}{3} + O(x^5)$

Conclusion:

The first four terms of the Maclaurin series for $\tan(x)$ are:

$\tan(x) \approx x + \frac{x^3}{3}$

Additional Questions:

1. How do we find the Maclaurin series for other trigonometric functions like $\sin(x)$ or $\cos(x)$?
2. What is the radius of convergence for the Maclaurin series of $\tan(x)$?
3. How would the Maclaurin series change if we were looking for a different function, such as $\arctan(x)$?
4. What are some common applications of Maclaurin series in physics or engineering?
5. How do you compute higher-order derivatives for more complicated functions?

Tip: Understanding how to derive and apply series expansions is useful for approximating functions, especially when dealing with small values of $x$ in practical applications.