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Klio2033 [76]
3 years ago
12

Find the second partial derivatives of the following functions

Mathematics
1 answer:
artcher [175]3 years ago
6 0

(a) <em>z</em> = 3<em>x</em> ² - 4<em>xy</em> + 15<em>y</em> ²

has first-order partial derivatives

∂<em>z</em>/∂<em>x</em> = 6<em>x</em> - 4<em>y</em>

∂<em>z</em>/∂<em>y</em> = -4<em>x</em> + 30<em>y</em>

and thus second-order partial derivatives

∂²<em>z</em>/∂<em>x</em> ² = 6

∂²<em>z</em>/∂<em>x</em>∂<em>y</em> = -4

∂²<em>z</em>/∂<em>y</em>∂<em>x</em> = -4

∂²<em>z</em>/∂<em>y</em> ² = 30

where ∂²<em>z</em>/∂<em>x</em>∂<em>y</em> = ∂/∂<em>x</em> [∂<em>z</em>/∂<em>y</em>] and ∂²<em>z</em>/∂<em>y</em>∂<em>x</em> = ∂/∂<em>y</em> [∂<em>z</em>/∂<em>x</em>].

(b) <em>z</em> = 4<em>x</em> <em>eʸ</em>

∂<em>z</em>/∂<em>x</em> = 4<em>eʸ</em>

∂<em>z</em>/∂<em>y</em> = 4<em>x</em> <em>eʸ</em>

∂²<em>z</em>/∂<em>x</em> ² = 0

∂²<em>z</em>/∂<em>x</em>∂<em>y</em> = 4<em>eʸ</em>

∂²<em>z</em>/∂<em>y</em>∂<em>x</em> = 4<em>eʸ</em>

∂²<em>z</em>/∂<em>y</em> ² = 4<em>x</em> <em>eʸ</em>

<em />

(c) <em>z</em> = 6<em>x</em> ln(<em>y</em>)

∂<em>z</em>/∂<em>x</em> = 6 ln(<em>y</em>)

∂<em>z</em>/∂<em>y</em> = 6<em>x</em>/<em>y</em>

∂²<em>z</em>/∂<em>x</em> ² = 0

∂²<em>z</em>/∂<em>x</em>∂<em>y</em> = 6/<em>y</em>

∂²<em>z</em>/∂<em>y</em>∂<em>x</em> = 6/<em>y</em>

∂²<em>z</em>/∂<em>y</em> ² = -6<em>x</em>/<em>y</em> ²

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