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# 15.5%20Ex%2050%20-%2056 - 694 C H A P T E R 15 D I F F E R...

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694 C H A P T E R 15 DIFFERENTIATION IN SEVERAL VARIABLES (ET CHAPTER 14) 50. Find a function f ( x , y , z ) such that f = z , 2 y , x . SOLUTION f ( x , y , z ) = xz + y 2 is a good choice. 51. Find a function f ( x , y ) such that f = y , x . SOLUTION We must find a function f ( x , y ) such that f = f x , f y = y , x That is, f x = y , f y = x We integrate the first equation with respect to x . Since y is treated as a constant, the constant of integration is a function of y . We get f ( x , y ) = y dx = yx + g ( y ) (1) We differentiate f with respect to y and substitute in the second equation. This gives f y = y ( yx + g ( y )) = x + g ( y ) Hence, x + g ( y ) = x g ( y ) = 0 g ( y ) = C Substituting in (1) gives f ( x , y ) = yx + C One of the solutions is f ( x , y ) = yx (obtained for C = 0). 52. Show that there does not exist a function f ( x , y ) such that f = y 2 , x . Hint: Use Clairaut’s Theorem f xy = f yx . SOLUTION Suppose that for some differentiable function f ( x , y ) , f = f x , f y = y 2 , x That is, f x = y 2 and f y = x . Therefore, f xy = y f x = y y 2 = 2 y and f yx = x f y

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