f � x � y � z � n f x y z for all � and we must show that f x � x � y � z � n 1

F ? x ? y ? z ? n f x y z for all ? and we must show

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f ( λ x , λ y , λ z ) = λ n f ( x , y , z ) for all λ , and we must show that f x ( λ x , λ y , λ z ) = λ n 1 f x ( x , y , z ) . We use the limit definition of f x . Since for all λ = 0, λ h 0 if and only if h 0, we get f x ( λ x , λ y , λ z ) = lim h 0 f ( λ x + λ h , λ y , λ z ) f ( λ x , λ y , λ z ) λ h = lim h 0 f ( λ ( x + h ), λ y , λ z ) f ( λ x , λ y , λ z ) λ h = lim h 0 λ n f ( x + h , y , z ) λ n f ( x , y , z ) λ h = lim h 0 λ n 1 f ( x + h , y , z ) λ n 1 f ( x , y , z ) h = λ n 1 lim h 0 f ( x + h , y , z ) f ( x , y , z ) h = λ n 1 f x ( x , y , z ) Alternatively, we prove this property using the Chain Rule. We use the Chain Rule to differentiate the following equality with respect to x : f ( λ x , λ y , λ z ) = λ n f ( x , y , z ) We get f x ( λ x , λ y , λ z ) · ∂( λ x ) x + f y ( λ x , λ y , λ z ) · ∂( λ y ) x + f z ( λ x , λ y , λ z ) · ∂( λ z ) x = λ n f x ( x , y , z ) Since ∂( λ y ) x = ∂( λ z ) x = 0 and ∂( λ x ) x = λ , we obtain for λ = 0, λ f x ( λ x , λ y , λ z ) = λ n f x ( x , y , z ) or f x ( λ x , λ y , λ z ) = λ n 1 f x ( x , y , z ) 39. Prove that if f ( x , y , z ) is homogeneous of degree n , then x f x + y f y + z f z = nf 9 Hint: Let F ( t ) = f ( tx , ty , tz ) and calculate F ( 1 ) using the Chain Rule. SOLUTION We use the Chain Rule to differentiate the function F ( t ) = f ( tx , ty , tz ) with respect to t . This gives F ( t ) = f x · ∂( tx ) t + f y · ∂( ty ) t + f z · ∂( tz ) t = x f x + y f y + z f z (1) On the other hand, since f is homogeneous of degree n , we have F ( t ) = f ( tx , ty , tz ) = t n f ( x , y , z ) Differentiating with respect to t we get F ( t ) = nt n 1 f ( x , y , z ) (2) By (1) and (2) we obtain x f x + y f y + z f z = nt n 1 f ( x , y , z ) Substituting t = 1 gives x f x + y f y + z f z = nf 40. Verify Eq. (9) for the functions in Exercise 37. SOLUTION Eq. (9) states that if f is homogeneous of degree n , then x f x + y f y + z f z = nf (a) f ( x , y , z ) = x 2 y + xyz . f is homogeneous of degree n = 3. The partial derivatives of f are f x = 2 xy + yz , f y = x 2 + xz , f z = xy Hence, x f x + y f y + z f z = x ( 2 xy + yz ) + y ( x 2 + xz ) + zxy = 3 x 2 y + 3 xyz = 3 ( x 2 y + xyz ) = 3 f ( x , y , z )
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