How do I do this the way I am supposed to? (without differentiating 9 times)
Any help would be greatly appreciated!
Thanks in advance!
Use the hint. The Taylor series for $\displaystyle \tan^{-1}x$ is
$\displaystyle
\tan^{-1}x = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots.
$
Subs in for $\displaystyle x \rightarrow \frac{x^3}{4}$. You really only need to consider powers up to $\displaystyle x^9$.
Ok I see how to get the summation of arctan(x^3 /4) but then I don't know what to do. Do I just write down the polynomials up to the x^9 power and do something there? What about f^(n) (a) x^n /n! formula ? What I was trying to do before my current attempt was to just do the summation of f^(n) (a) x^n /n! and to replace f^(n) with it's appropriate algebraic expression but it was tough to compute the second derivative and on.
Sorry for being stupid
So you're OK with this, correct?
$\displaystyle f(x) = \arctan \left(\frac{x^3}{4}\right) = \frac{x^3}{4} - \frac{x^9}{4^3 \cdot 3} + \frac{x^{15}}{4^5 \cdot 5} - \frac{x^{21}}{4^7 \cdot 7} + \cdots$
Now, consider what happens when you take 9 derivatives of this expression.
The first term $\displaystyle \frac{x^3}{4}$ will eventually become zero (beyond 3 derivatives).
The second term $\displaystyle -\frac{x^9}{4^3 \cdot 3}$ will become a constant. What constant does it become?
The remaining terms will turn into the form $\displaystyle C \cdot x^n$ where $\displaystyle C$ is a constant and $\displaystyle n$ is a positive integer. When you substitute $\displaystyle x=0$ into a term of this form, does it make sense that these terms will all become zero?
Therefore, there is only one nonzero term, which is the second term.