Find All the Second Order Partial Derivatives of the Given $f ( x , y ) = \cos \left( x y ^ { 2 } \right)$

question 300

Multiple Choice

Find all the second order partial derivatives of the given function.
- $f ( x , y ) = \cos \left( x y ^ { 2 } \right)$

• A) \(\frac { \partial ^ { 2 } f } { \partial x ^ { 2 } } = - y ^ { 4 } \cos x y ^ { 2 } ; \frac { \partial ^ { 2 } f } { \partial y ^ { 2 } } = - 2 x \left[ 2 x y ^ { 2 } \cos \left( x y ^ { 2 } \right) + \sin \left( x y ^ { 2 } \right) \right] ; \frac { \partial ^ { 2 } f } { \partial y \partial x } = \frac { \partial ^ { 2 } f } { \partial x \partial y } =\) \(- 2 y \left[ x y ^ { 2 } \cos \left( x y ^ { 2 } \right) + \sin \left( x y ^ { 2 } \right) \right] ;\)
• B) \(\frac { \partial ^ { 2 } f } { \partial x ^ { 2 } } = y ^ { 2 } \sin x y ^ { 2 } ; \frac { \partial ^ { 2 } f } { \partial y ^ { 2 } } = 2 \left[ 2 y ^ { 2 } \cos \left( x y ^ { 2 } \right) - \sin \left( x y ^ { 2 } \right) \right] ; \frac { \partial ^ { 2 } f } { \partial y \partial x } = \frac { \partial ^ { 2 } f } { \partial x \partial y } = 2 y \left[ y ^ { 2 } \cos \left( x y ^ { 2 } \right) - \sin \left( x y ^ { 2 } \right) \right]\)
• C) \(\frac { \partial ^ { 2 } \mathrm { f } } { \partial \mathrm { x } ^ { 2 } } = - \mathrm { y } ^ { 2 } \sin x \mathrm { y } ^ { 2 } ; \frac { \partial ^ { 2 } \mathrm { f } } { \partial \mathrm { y } ^ { 2 } } = 2 \left[ \sin \left( \mathrm { xy } ^ { 2 } \right) - 2 \mathrm { y } ^ { 2 } \cos \left( x \mathrm { y } ^ { 2 } \right) \right] ; \frac { \partial ^ { 2 } \mathrm { f } } { \partial \mathrm { y } \partial \mathrm { x } } = \frac { \partial ^ { 2 } \mathrm { f } } { \partial \mathrm { x } \partial \mathrm { y } } = 2 \mathrm { y } \left[ \sin \left( x \mathrm { y } ^ { 2 } \right) - \mathrm { y } ^ { 2 } \cos \left( x \mathrm { y } ^ { 2 } \right) \right]\)
• D) \(\frac { \partial ^ { 2 } f } { \partial x ^ { 2 } } = - y ^ { 2 } \sin x y ^ { 2 } ; \frac { \partial ^ { 2 } f } { \partial y ^ { 2 } } = 2 y \left[ 2 y ^ { 2 } \cos \left( x y ^ { 2 } \right) - \sin \left( x y ^ { 2 } \right) \right] ; \frac { \partial ^ { 2 } f } { \partial y \partial x } = \frac { \partial ^ { 2 } f } { \partial x \partial y } = 2 \left[ y ^ { 2 } \cos \left( x y ^ { 2 } \right) - \sin \left( x y ^ { 2 } \right) \right]\)

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