Science:Math Exam Resources/Courses/MATH101/April 2011/Question 01 (j)/Solution 1

We could compute all of the derivatives by hand, but this would be a long and arduous process. This is not an efficient use of exam time! As such, we use the known series of $\sin (x)$ to note that

$\sin (x^3)=x^3-\frac{x^9}{3!}+\frac{x^{15}}{5!}-\cdots$

It follows then that

$x^3\sin (x^3)=x^3(x^3-\frac{x^9}{3!}+\frac{x^{15}}{5!}-\cdots )=x^6-\frac{x^{12}}{3!}+\frac{x^{18}}{5!}-\cdots$

and so the first three nonzero terms of the Maclurian series are

$x^6-\frac{x^{12}}{3!}+\frac{x^{18}}{5!}$