Finding the 5th order taylor polynomial of a function without brute force I am currently stuck on this 5th order taylor polynomial question for almost an hour now, it's quite difficult if you brute force it but there was a hint given today by my tutor that there is an easier way to solve for the 5th taylor polynomial of this function below if you do some substitutions. Should I be substituting $x^2$ for $x$ and then find the 5th taylor polynomial and then substituting $x^2$ back in at the end or am I completely off with this suggestion?:$$f(x) = \frac{\sin(x^2)}{x^2}$$
 A: HINT Yes, substitute $u=x^2,$ so you have $\sin(u)/u.$ Then find the (easier) expansion of $\sin(u)$ out a couple terms and divide through by $u$. Then sub back in $x^2.$   
A: Building on the hint $u=x^2$.
$\sin(u)=u-\frac{u^3}{3!}+\frac{u^5}{5!}$
$\frac{\sin(u)}{u}=1-\frac{u^2}{3!}+\frac{u^4}{5!}$
Now back substituting in $x^2$
$\frac{\sin(x^2)}{x^2}=1-\frac{x^4}{3!}+\frac{x^8}{5!}+...$
Edit
Using $x=x^2$ as the substitution as op had.
$\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}+...$
$\frac{\sin(x)}{x}=1-\frac{x^2}{3!}+\frac{x^4}{5!}+...$
Now back substituting $x^2$
$\frac{\sin(x^2)}{x^2}=1-\frac{(x^2)^2}{3!}+\frac{(x^2)^4}{5!}...$
It doesn't matter what you pick as your substitution.  It may get confusing if you use the substitution $x=x^2$.
A: More precisely, you should be substituting $x^2$ for $t$ and develop until $x^5$, i.e. $t^{5/2}$ !?
The fact is that, as this substitution is possible, the original function is even and has no odd powers in its development. So the fifth and fourth order developments are the same and it suffices to develop up to $t^2$.
Direct attack doesn't look so monstruous,
$$g(t)=\frac{\sin(t)}t,\\g'(t)=-\frac{\sin t}{t^2}+\frac{\cos t}t,\\
g''(t)=2\frac{\sin t}{t^3}-2\frac{\cos t}{t^2}-\frac{\sin t}t,$$ but you still have to take limits at $0$, for instance by L'Hospital, requiring further differentiations (though by parity considerations, we can anticipate $g'(0)=0$).
Certainly, developing $\sin t$ and dividing by $t$ is the most efficient approach, immediately giving 
$$1-\frac{t^2}{3!}=1-\frac{x^4}6.$$

We can also attempt without the $t$ substitution, and develop $\sin(x^2)$ up to degree $6$:
$$h(x)=\sin(x^2),\\
h'(x)=2x\cos(x^2),\\
h''(x)=-4x^2\sin(x^2)+2\cos(x^2),\\
h'''(x)=-8x^3\cos(x^2)-12x\sin(x^2),\\
h''''(x)=16x^4\sin(x^2)-48x^2\cos(x^2)-12\sin(x^2),\\
h'''''(x)=32x^5\cos(x^2)+160x^3\sin(x^2)-120x\cos(x^2),\\
h''''''(x)=-64x^6\sin(x^2)+480x^4\cos(x^2)+720x^2\sin(x^2)-120\cos(x^2).
$$
This gives us
$$h(x)=\frac2{2!}x^2-\frac{120}{6!}x^6$$
and in the end,
$$1-\frac{x^4}6.$$
