Calculus problem? (High-amount of advanced math equation included)

I was on the internet randomly searching for derivative worksheets to practice AP Calculus AB when I came across this:

$f\left(t\right)=\sin ^5\left(e^{t^2}-\cos \left(t\right)\right)$

I used the triple composite function of chain rule two times to solve this:

$\frac{d}{dt}\left[\sin ^5\left(e^{t^2}-\cos \left(t\right)\right)\right]$

Apply advanced chain rule:

Set:

$u\left(t\right)=e^{t^2}-\cos \left(t\right)$

$v\left(t\right)=\sin \left(t\right)$

$w(t)=t^5$

The function can be rewritten using these three function as:

$f(t)=w(v(u(t)))$

$\frac{d}{dt}\left[w\left(v\left(u\left(t\right)\right)\right)\right]=w'\left(v\left(u\left(t\right)\right)\right).v'\left(u\left(t\right)\right).u'\left(t\right)$

Unsubstitute $u(t), v(t),$and $w(t)$, I got:

$\frac{d}{dt}\left[\sin ^5\left(e^{t^2}-\cos \left(t\right)\right)\right]$

$=4\left(\sin ^4\left(e^{t^2}-\cos \left(t\right)\right)\right).\cos \left(e^{t^2}-\cos \left(t\right)\right).\frac{d}{dx}\left[e^{t^2}-\cos \left(t\right)\right]$

Applying chain rule for the second time on $e^{t^2}$

$x(t)=t^2$

$y(t)=e^t$

The function $e^{t^2}$ can be rewritten as:

$y(x(t))$

Apply regular chain rule:

$\frac{d}{dt}\left[y\left(x\left(t\right)\right)\right]=y'\left(x\left(t\right)\right).x'\left(t\right)$

Unsubstitute $x(t)$ and $y(t)$, I got

$\frac{d}{dt}\left[e^{t^2}\right]=e^{t^2}.2t=2te^{t^2}$

$\frac{d}{dt}\left[-\cos \left(t\right)\right]=\sin \left(t\right)$

Place the solved integral back:

$\frac{d}{dt}\left[\sin ^5\left(e^{t^2}-\cos \left(t\right)\right)\right]$

$=4\left(\sin ^4\left(e^{t^2}-\cos \left(t\right)\right)\right).\cos \left(e^{t^2}-\cos \left(t\right)\right).\left(2te^{t^2}+\sin \left(t\right)\right)$

Yet when I compare the correct answer to my answer, my answer seems to be constantly about 3/5 times the correct answer's value, I stared eye-to-eye to my process for three hours but can't find any problems, can I get anyone's help? :P

People who can help me will be highly-appreciated :D