# Multivariable Instantaneous rate of change clarification When you are computing the instantaneous rate of change for f(x,y) what do you take the derivative with respect to? for example, for f(x,y)=(sin(pi^x)cos(pi^y),ye^xy,x^2+y^3) If I was to find the instantaneous rate of change for all 3 of these functions going through (1,2) with the velocity vector (3,-2) would I just take d/dx of all of the functions at (3,-2)?

Multivariable Instantaneous rate of change clarification
When you are computing the instantaneous rate of change for f(x,y) what do you take the derivative with respect to?
for example, for
f(x,y)=(sin(πx)cos(πy),yexy,x2+y3)
If I was to find the instantaneous rate of change for all 3 of these functions going through (1,2) with the velocity vector (3,-2) would I just take ddx of all of the functions at (3,-2)?
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If you want the change in f as both x and y are changing, then you are going to need something that relates x to y. (a parametric curve perhaps, or a direction vector).
Without that, you can evaluate the sentivity of f to changes in x this is the partial derivative. $\frac{\mathrm{\partial }f}{\mathrm{\partial }x}$
And there is also a partial derivative with respect to y, i.e. $\frac{\mathrm{\partial }f}{\mathrm{\partial }y}$
If you have the partials, and a parametric curve then you can find the "total derivative"
$\frac{df}{dt}=\frac{\mathrm{\partial }f}{\mathrm{\partial }x}\frac{dx}{dt}+\frac{\mathrm{\partial }f}{\mathrm{\partial }y}\frac{dy}{dt}$ and these will correspond to the same derivative you would get if you subsituted x(t),y(t) into your function.
Since f is a vector, you will get a derivatives for each of the i,j,k components.