Unformatted text preview: v t =0 [ f ( −→ x + t −→ e j )] or f x ( x,y,z ) := ∂f ∂x ( x,y,z ) = lim t → 1 t [ f ( x + t,y,z ) − f ( x,y,z )] f y ( x,y,z ) := ∂f ∂y ( x,y,z ) = lim t → 1 t [ f ( x,y + t,z ) − f ( x,y,z )] f z ( x,y,z ) := ∂f ∂z ( x,y,z ) = lim t → 1 t [ f ( x,y,z + t ) − f ( x,y,z )] . In practice, partial derivatives are easy to calculate: we just diFerentiate, treating all but one of the variables as a constant. ±or example, if f ( x,y ) = x 2 y + 3 x + 4 y then ∂f ∂x , the partial with respect to x , is obtained by treating y as the name of some constant: f x ( x,y ) := ∂f ∂x ( x,y ) = 2 xy + 3 6 The symbol ∂f ∂x j is pronounced as if the ∂ ’s were d ’s....
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 Fall '08
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 Calculus, Derivative

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