Say I have the function $g(t) = f(4 \sin 2t , 5 \cos 2t)$, how do I find $g'(t)$? More specifically, if the formulaic method of getting $g'(t)$ is deriving the various partial derivatives in this manner,
$\frac{dg}{dx}*\frac{dx}{dt}$+$\frac{dg}{dy}*\frac{dy}{dt}$
how would one find $\frac{dg}{dx}$ or $\frac{dg}{dy}$?
EDIT: As @Arthur stated below, the $\frac{dg}{dx}$ and $\frac{dg}{dy}$ in the formula above should be replaced with $\frac{df}{dx}$ and $\frac{df}{dy}$, respectively.
$g$ is a function of a single variable, so there is no $\frac{\partial g}{\partial x}$ or $\frac{\partial g}{\partial y}$. There is only $\frac{dg}{dt}$.
$f$, on the other hand, is a different story. It is a function of two variables, so the multivariable chain rule applies there.
We have $$ g(t)=f(x(t), y(t))\\ x(t)=4\sin 2t\\ y(t)=5\cos 2t $$ Now differentiating the first line by $t$ on both sides, and using the multivariable chain rule on $f$, we get $$ \frac{dg}{dt}=\frac{d}{dt}f(x,y)\\ =\frac{\partial f}{\partial x}\cdot \frac{dx}{dt}+\frac{\partial f}{\partial y}\cdot\frac{dy}{dt}\\ =\frac{\partial f}{\partial x}\cdot 8\cos 2t-\frac{\partial f}{\partial y}\cdot10\sin 2t\\ $$ We can't do any more than that without knowing what the partial derivatives of $f$ are.