Answer
$3x^4-\dfrac{27x^{12}}{3}+\dfrac{243x^{20}}{5}-\dfrac{2187x^{28}}{7}$
Work Step by Step
Since, we know that the Maclaurin Series for $\tan^{-1} x$ is defined as:
$ \tan^{-1} x=\Sigma_{n=0}^\infty (-1)^n\dfrac{x^{2n+1}}{(2n+1)!}=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+...$
Now, $ \tan^{-1} (3x^4)=(3x^4)-\dfrac{(3x^4)^3}{3!}+\dfrac{(3x^4)^5}{5!}-\dfrac{(3x^4)^7}{7!}+...$
or, $=3x^4-\dfrac{27x^{12}}{3}+\dfrac{243x^{20}}{5}-\dfrac{2187x^{28}}{7}$
