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
to note that
![{\displaystyle \sin(x^{3})=x^{3}-{\frac {x^{9}}{3!}}+{\frac {x^{15}}{5!}}-\cdots }](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/a8066dd8c9d3e4a3e6745e8cfa1da7d84d46545a)
It follows then that
![{\displaystyle 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 }](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/e4ab18f719802ec52fff8ef144854cf21d561a4a)
and so the first three nonzero terms of the Maclurian series are
![{\displaystyle x^{6}-{\frac {x^{12}}{3!}}+{\frac {x^{18}}{5!}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/d086fbb6472466708611b660dbcd998c81137e66)