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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?

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Please consider learning some $\LaTeX$ to format your questions. Make sure I didn't alter the meaning of it. Also, $\frac{\partial F}{\partial x}$ and $\frac{\partial F}{\partial z}$ are wrong. –  Pragabhava Oct 24 '12 at 18:34
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Go to FAQ to learn a little about LaTeX. Now, are you sure that F is correctly defined? Because you have $\,F=2x^2-yz+x^2\,$...This looks weird. –  DonAntonio Oct 24 '12 at 18:42
    
@DonAntonio: I think $F(x,y,z)=2x^2-xz^2-yz$ but $x,y,z$ seems to be printed wrong. –  B. S. Oct 24 '12 at 18:57
    
@ViviAluvieriti: Are you sure all $x,y,z$ are defined correctly with respect to $t$? Can you check them again? –  B. S. Oct 24 '12 at 19:21
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@DonAntonio If you take $z = 3 e^{-t}$ you get the desired result. There must be a typo on the book :) –  Pragabhava Oct 24 '12 at 23:25
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2 Answers 2

$$\frac{\partial F}{\partial t}=\frac{\partial F}{\partial x}\frac{dx}{dt}+\frac{\partial F}{\partial y}\frac{dy}{dt}+\frac{\partial F}{\partial z}\frac{dz}{dt}=$$

$$=(4x+z^2)\cdot 2\cos t-z(2t-1)+(2xz-y)\cdot 0$$

You'll now to substitute:

$$t=0\Longrightarrow\,x=0\,,\,y=1\,,\,z=3\,e^{-1}$$

The final result is, if I'm not wrong,

$$9\,e^{-2}+3\,e^{-1}$$

which has nothing to do with $\,24\,$, so either the book (which one, btw?) has a mistake or you miscopied the exercise.

Ps. Please do you check that according to what you wrote $\,z\,$ is independent of $\,t\,$...

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You need to use implicit differentiation as one of your tags suggests.

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I think you formula for $dF/dt$ is wrong. Do you know that a total derivative is? –  glebovg Oct 24 '12 at 18:44
    
You are missing $\partial F/\partial t$ in your formula for the total derivative. –  glebovg Oct 24 '12 at 20:09
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