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## Homework Statement

Suppose the differentiable function f(x,y,z) has the partial derivatives f

_{x}(1,0,1) = 4, f

_{y}(1,0,1) = 1 and f

_{z}(1,0,1) = 0. Find g'(0) if g(t) = f(t

^{2}+ 1, t

^{2}-t, t+1).

## Homework Equations

## The Attempt at a Solution

Ok I'm given the solution for this and I'm trying to work through it but I'm confused.

I understand we have g(t) = f(x(t),y(t),z(t)) where x = t

^{2}+1; y = t

^{2}-t; z = t+1.

So i thought I would just use the chain rule to find g'(t) and plug in 0 for t. But I check the solutions sheet and he uses a very different (easier) method that I don't understand. ( I just realized that we aren't given a function f in terms of x y or z so thats why this wouldn't work and we need another method. Still I don't understand this other method.)

First he says when t=0 (x,y,z)|

_{t=0}= (1,0,0) (I think he may have made a mistake, shouldn't it be (1,0,1) ?) And then he gets a point P = (1,-1,0) No idea were this point comes from...

Next he sets up the chain rule equation: [tex]\frac{dg}{dt} = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt} + \frac{\partial f}{\partial z} \frac{dz}{dt}[/tex]

Now I think he somehow uses that point and the partial derivatives given above to solve the equation and he gets: = 1*0 + 1*(-1) + (0*1) = -1 (depending on whether P was correct above may change the answer here I think)

So could someone please explain the method used here?