I have a question. How in general would one differentiate a composite function like $F(x,y,z)=2x^2-yz+xz^2$ where $x=2\sin t$ , $y=t^2-t+1$ , and $z = 3e^{-1}$ ? I want to find the value of $\frac{dF}{dt}$ evaluated at $t=0$ and I don't know how. Can someone please walk me through this?
I tried a couple of things, including chain rules and jacobians. I know that $\frac{dF}{dt}$ should equal $\frac{\partial F}{\partial x} \frac{dx}{dt} + \frac{\partial F}{\partial y} \frac{dy}{dt} + \frac{\partial F}{\partial z} \frac{dz}{dt}$ but for some reason this doesn't work, or I am doing something wrong. I start out by differentiating to get $\frac{\partial F}{\partial x}=4x+z^2$, $\frac{\partial F}{\partial y}= -z$, $\frac{\partial F}{\partial z} = 2xz-y$, $\frac{dz}{dt}=0$, $\frac{dx}{dt}=2\cos t$, $\frac{dy}{dt}=2t-1$ but this doesn't match the answer, which my book says is $24$.
How do they get this, and where is my error? Thanks.
Update:
What I get is as follows: $F(x,y,z)=2x^2-yz+xz^2$, $\frac{\partial F}{\partial t}=\frac{\partial F}{\partial x} \frac{dx}{dt} + \frac{\partial F}{\partial y} \frac{dy}{dt} + \frac{\partial F}{\partial z} \frac{dz}{dt}$,$\frac{\partial F}{\partial t}=(4x+z^2)(2cos(t))-z(2t-1)$ Which for $t=0$ gives $x=0$ and $\left. \frac{\partial F}{\partial t} \right|_{t=0} = 2z^2+z=9e^{-2}+3e^{-1}$ which clearly isn't $24$ so I must be doing something completely wrong.
Edit: I want to rephrase the question. Since everyone else I have talked to thinks there was an error in the book, does everyone here agree?