Originally Posted by

**Diemo** What? So you let u be 2t. Then you have $\displaystyle \frac{dy}{dt}=\frac{dy}{du}\frac{du}{dt}$. Now, y=sin(u), u=2t.

Hence

$\displaystyle \frac{du}{dt}=2$ and $\displaystyle \frac{dy}{du} = cos(u)$

Hence, $\displaystyle \frac{dy}{dt}= \frac{dy}{du}\frac{du}{dt}=cos(u)*2=2cos(u)=2*cos( 2t)$, where I have resubstituted the u=2t at the end. Now do the same for your second equation, only now you let y= u^2, u = sec(3t), and to work out what the derivitive of sec(3t) is you use the chain rule again, letting u=3t, y=sec(u).