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

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

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

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