Engineering Calculus Notes 246

# Engineering Calculus Notes 246 - v t =0 f −→ x t −→...

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234 CHAPTER 3. REAL-VALUED FUNCTIONS: DIFFERENTIATION Partial Derivatives Equation ( 3.6 ), combined with Remark 3.2.1 , gives us a way of expressing the differential d −→ x 0 f ( −→ v ) as a homogeneous polynomial in the components of −→ v . The quantity given by Equation ( 3.6 ) when −→ v = −→ e j is an element of the standard basis for R 3 , is called a partial derivative . It corresponds to moving through −→ x 0 parallel to one of the coordinate axes with unit speed—that is, the motion parametrized by −→ p j ( t ) = −→ x 0 + t −→ e j : Definition 3.3.3. The j th partial derivative (or partial with respect to x j ) of a function f ( x 1 ,x 2 ,x 3 ) of three variables at −→ x = −→ x 0 is the derivative 6 (if it exists) of the function ( f −→ p j )( t ) obtained by fixing all variables except the j th at their values at −→ x 0 , and letting x j vary: f x j ( −→ x 0 ) := ∂f ∂x j ( −→ x 0 ) = d dt vextendsingle vextendsingle vextendsingle vextendsingle t =0 [ f ( −→ p j ( t ))] = d dt vextendsingle
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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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