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**Showcase_22** That's what I thought as well. Here's how my substitution goes:

$\displaystyle t(-16t^2 sin(t^2)+12t^2 cos (t^2))-(8tcos(t^2)+6tsin(t^2))+4t^3 (4sin (t^2)-3cos(t^2))$

$\displaystyle 4t^3(-4sin(t^2)+3cos(t^2))-2t(4cos(t^2)+3sin(t^2))+4t^3(4sin(t^2)-3cos(t^2))$

$\displaystyle 4t^3(-4sin(t^2)+3cos(t^2)+4sin(t^2)-3cos(t^2))-2t(4cos(t^2)+3sin(t^2))$

$\displaystyle =-2t(4cos(t^2)+3sin(t^2))=-\frac{dy}{dt}$

I did it a different way to how I did before and I now have this answer. I can't see what i'm doing wrong or if it's a typo on the homework sheet.