Explanation Given Information: The provided function is z = f ( x , y ) = 5 x − 9 y 2 + 2 ( x y + 1 ) 5 . Formula used: For a function f ( x , y ) , the partial derivative of f with respect to x is calculated by taking the derivative of f ( x , y ) with respect to x and keeping the other variable y constant. The partial derivative of f with respect to x is denoted by f x and the partial derivative of f with respect to y is denoted by f y . For a function z ( x , y ) , the second partial derivative, (1) When both derivatives are taken with respect to x is z x x = ∂ 2 z ∂ x 2 = ∂ ∂ x ( ∂ z ∂ x ) . (2) When both derivatives are taken with respect to y is z y y = ∂ 2 z ∂ y 2 = ∂ ∂ y ( ∂ z ∂ y ) . (3) When first derivative is taken with respect to x and second derivative is taken with respect to y is z x y = ∂ 2 z ∂ y ∂ x = ∂ ∂ y ( ∂ z ∂ x ) . (4) When first derivative is taken with respect to y and second derivative is taken with respect to x is z y x = ∂ 2 z ∂ x ∂ y = ∂ ∂ x ( ∂ z ∂ y ) . Power of x rule for a real number n is such that, if f ( x ) = x n then f ′ ( x ) = n x n − 1 . Constant function rule for a constant c is such that, if f ( x ) = c then f ′ ( x ) = 0 . Chain rule for function f ( x ) = u ( v ( x ) ) is f ′ ( x ) = u ′ ( v ( x ) ) ⋅ v ′ ( x ) . Coefficient rule for a constant c is such that, if f ( x ) = c ⋅ u ( x ) , where u ( x ) is a differentiable function of x , then f ′ ( x ) = c ⋅ u ′ ( x ) . Calculation: Consider the function, z = f ( x , y ) = 5 x − 9 y 2 + 2 ( x y + 1 ) 5 . Recall that, for a function f ( x , y ) , the partial derivative of f with respect to x is calculated by taking the derivative of f ( x , y ) with respect to x and keeping the other variable y constant. Use the power of x rule for derivatives, the constant function rule, the chain rule and the coefficient rule, z x = ∂ z ∂ x = 5 + 2 ( 5 ( x y + 1 ) 4 ) ⋅ ( y ) = 5 + 10 y ( x y + 1 ) 4 And, z y = ∂ z ∂ y = − 9 ( 2 y ) + 2 ( 5 ( x y + 1 ) 4 ) ⋅ ( x ) = − 18 y + 10 x ( x y + 1 ) 4 Hence, the first partial derivatives of the function z = f ( x , y ) = 5 x − 9 y 2 + 2 ( x y + 1 ) 5 are z x = 5 + 10 y ( x y + 1 ) 4 and z y = − 18 y + 10 x ( x y + 1 ) 4 . Recall that, for a function z ( x , y ) , the second partial derivative, when both derivatives are taken with respect to x is z x x = ∂ 2 z ∂ x 2 = ∂ ∂ x ( ∂ z ∂ x ) . Use the power of x rule for derivatives, the constant function rule and the coefficient rule. z x x = ∂ 2 z ∂ x 2 = ∂ ∂ x ( ∂ z ∂ x ) = ∂ ∂ x ( 5 + 2 ( 5 ( x y + 1 ) 4 ) ⋅ ( y ) ) = ∂ ∂ x ( 5 + 10 y ( x y + 1 ) 4 ) Simplify it further, z x x = 10 y ( 4 ( x y + 1 ) 3 ( y ) ) = 40 y 2 ( x y + 1 ) 3 Recall that, for a function z ( x , y ) , the second partial derivative, when first derivative is taken with respect to x and second derivative is taken with respect to y is z x y = ∂ 2 z ∂ y ∂ x = ∂ ∂ y ( ∂ z ∂ x )

11th Edition

ISBN: 9781305713864

Author: HARSHBARGER

Publisher: CENGAGE L

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Question

Chapter 14, Problem 2T

To determine

To calculate: All the first and second partial derivatives of z=f(x,y)=5x−9y2+2(xy+1)5.

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Chapter 14, Problem 1T

Chapter 14, Problem 3T

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