<s>[INST] Differentiation and Rate of Change Im slaving over this one
using rate of change and Differentiation
I tried the chain rule but that is not the route as it got passed back
Question: find rate of change v if 
V = x^3 * y^2
where
x=cos3t and y =sin3t
which gives
v = cos^3(3t) * sin^2(3t)
This is what i have so far 
dv/dx
3-3Sin3t^2 x Sin3t^2
-9sin3t^2 x Sin3t^2
-9Sin3t^4
dv/dy
cos3t^3 x 2x3Cos3t 
Cos3t^3 x 6Cos3t
6Cos3t^4
so
v = -9Sin3t^4 + 6Cos3t^4
but get the feeling that the "product rule" changes my calculations further.
im a bit stuck, can any one give some advice or confirm my calcs ?
Thanks
 [/INST] Since $x=\cos 3t$ and $y = \sin 3t$, by chain rule we have:
$$
\dfrac{dx}{dt} = -3 \sin 3t \qquad \text{and} \qquad \dfrac{dy}{dt} = 3 \cos 3t
$$
Hence, by product rule, we have:
$$ \begin{align*}
\dfrac{dV}{dt}
&= \left(3x^2 \dfrac{dx}{dt}\right)(y^2) + (x^3)\left(2y \dfrac{dy}{dt}\right) \\
&= (3\cos^2 3t) (-3 \sin 3t)(\sin^2 3t) + (\cos^3 3t)(2 \sin 3t) (3 \cos 3t) \\
&= -9\cos^2 3t\sin^2 3t\sin 3t + 6\cos^3 3t\cos 3t\sin 3t \\
\end{align*} $$
</s>