If $x=t^2\sin3t$ and $y=t^2\cos3t$, find $\frac{dy}{dx}$ in terms of $t$ If $x=t^2\sin3t$ and $y=t^2\cos3t$, find $\frac{dy}{dx}$ in terms of $t$. This is how I tried solving it:
$$
\frac{dx}{dt} = 2t\sin3t + 3t^2\cos3t \\
\frac{dy}{dt} = 2t\cos3t - 3t^2\sin3t \\
\frac{dy}{dx} = \frac{2t\cos3t - 3t^2\sin3t}{2t\sin3t + 3t^2\cos3t} 
$$
But the answer listed is: 
$$
\frac{2-3t\tan3t}{2\tan3t+3t}
$$
Is my answer incorrect, or can I simplify it even more?

You are on the right trace. Just divide $t\cos 3t$ in the numerator and denominator of  $\frac{dy}{dx}$.
