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

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: $$ \begin{align*} \dfrac{dV}{dt} &= \left(3x^2 \dfrac{dx}{dt}\right)(y^2) + (x^3)\left(2y \dfrac{dy}{dt}\right) \\\ &= (3\cos^2 3t) (-3 \sin 3t)(\sin^2 3t) + (\cos^3 3t)(2 \sin 3t) (3 \cos 3t) \\\ &= -9\cos^2 3t\sin^2 3t\sin 3t + 6\cos^3 3t\cos 3t\sin 3t \\\ \end{align*} $$