Quote by Mark44
Not quite. You are confusing the derivatives of x(t), y(t), and z(t) with the value of each derivative at a specific value of t. IOW, x'(t) = 2t, but x'(0) = 0, and similarly for y'(t) vs. y'(0) and z'(t) vs. z'(0). In each case, the deriviative is a function, while the derivative evaluated at a particular value of t is just a number. x'(t) is different from x'(0), y'(t) is different from y'(0), and z'(t) is different from z'(0).

Isn't this what I did? I think I may have messed up my notation a bit but I have x'(t) = 2t then x'(0) = 0; y'(t) = 2t1 then y'(0) = 1; z'(t) = 1 then z'(0) = 1;
Quote by Mark44
The same difference exists between g'(t) and g'(0). Using the chain rule you can write g'(t) = f_{x}(x,y,z) * x'(t) + f_{y}(x,y,z) * y'(t) + f_{z}(x,y,z) * z'(t). Since you aren't given any details about f(x, y, z), it's not possible to calculate the functions f_{x}(x, y, z), f_{y}(x, y, z), and f_{z}(x, y, z). You are, however given the values of these partials at a particular point (1, 0, 1) that corresponds to t = 0.

So here since the point (1,0,1) corresponds to g(t) when t is 0 we can use the value of the partial at (1,0,1) as the value of f
_{x}(x,y,z)? Which would mean at (1,0,1) f
_{x} = 4; f
_{y} = 1; f
_{z} = 0.
Quote by Mark44
g'(0) = f_{x}(1,0,1) * x'(0) + f_{y}(1,0,1) * y'(0) + f_{z}(1,0,1) * z'(0). The only reason I have used the subscript notation (as opposed to the Leibniz notation you used) for the partial derivatives is that it's a little easier to explicitly show that they are to be evaluated at a particular point.

So if everything above is correct that gives me: 4*0 + 1*(1) + 0*1 = 1.
Also, does that mean (if this is all correct) this method can only be used if the point the partials are evaluated at and the value of t are directly related as in this example, correct?