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Question
If f(x, y) = 3x2 + 4y3 + 6xy - x2y3 + 7, then show that fyy (1,1) = 18.
Solution
Given f(x, y) = 3x2 + 4y3 + 6xy - x2y3 + 7
Differentiating partially with respect to 'y' we get,
`"f"_"y" = (del"f")/(del"y") = 0 + 12y^2 + 6x(1) - x^2(3y^2) + 0`
= 12y2 + 6x - 3x2y2
Differentiating again partially with respect to 'y' we get,
`"f"_"yy" = (del^2"f")/(del"y"^2) = 24y + 0 - 3x^2 (2y) = 24y - 6x^2y`
∴ fyy (1, 1) = 24(1) - 6(1)2(1) = 24 - 6 = 18
Hence proved.
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