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Calculate ∂2f∂x2\frac { \partial ^ { 2 } f } { \partial x ^ { 2 } }

Question 15

Multiple Choice

Calculate ∂2f∂x2\frac { \partial ^ { 2 } f } { \partial x ^ { 2 } } , ∂2f∂yx\frac { \partial ^ { 2 } f } { \partial y x } when defined. f(x,y) =4x0.9y0.7f ( x , y ) = 4 x ^ { 0.9 } y ^ { 0.7 }


A) ∂2f∂x2=−0.36y0.7x−1.1\frac { \partial ^ { 2 } f } { \partial x ^ { 2 } } = - 0.36 y ^ { 0.7 } x ^ { - 1.1 } , ∂2f∂yx=−2.52y0.3x0.1\frac { \partial ^ { 2 } f } { \partial y x } = - 2.52 y ^ { 0.3 } x ^ { 0.1 }
B) ∂2f∂x2=−3.6y0.7x−0.1\frac { \partial ^ { 2 } f } { \partial x ^ { 2 } } = - 3.6 y ^ { 0.7 } x ^ { - 0.1 } , ∂2f∂yx=2.8y−0.3x0.9\frac { \partial ^ { 2 } f } { \partial y x } = 2.8 y ^ { - 0.3 } x ^ { 0.9 }
C) ∂2f∂x2=−3.6y0.7x−1.1\frac { \partial ^ { 2 } f } { \partial x ^ { 2 } } = - 3.6 y ^ { 0.7 } x ^ { - 1.1 } , ∂2f∂yx=3.6y−0.3x−0.1\frac { \partial ^ { 2 } f } { \partial y x } = 3.6 y ^ { - 0.3 } x ^ { - 0.1 }
D) ∂2f∂x2=−0.36y0.7x−1.1\frac { \partial ^ { 2 } f } { \partial x ^ { 2 } } = - 0.36 y ^ { 0.7 } x ^ { - 1.1 } , ∂2f∂yx=2.52y−0.3x−0.1\frac { \partial ^ { 2 } f } { \partial y x } = 2.52 y ^ { - 0.3 } x ^ { - 0.1 }
E) ∂2f∂x2=−4y−0.3x−1.1\frac { \partial ^ { 2 } f } { \partial x ^ { 2 } } = - 4 y ^ { - 0.3 } x ^ { - 1.1 } , ∂2f∂yx=4y−1.3x−0.1\frac { \partial ^ { 2 } f } { \partial y x } = 4 y ^ { - 1.3 } x ^ { - 0.1 }

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