AKG
Nov10-04, 01:28 AM
Find the (first-order) partial derivatives of the following function (where g : \mathbb{R} \to \mathbb{R} is continuous):
f(x, y) = \int _a ^{\int _b ^y g} g.
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I got:
D_1f(x, y) = 0
D_2f(x, y) = \frac{\partial f}{\partial y}
= \frac{\partial }{\partial y} \int _a ^{\int _b ^y g}g
Let G be the antiderivative of g:
= \frac{\partial }{\partial y}\left ( G \left (\int _b ^y g\right ) - G(a) \right )
= \frac{\partial }{\partial y} G \left (\int _b ^y g\right )
= \left ( \frac{\partial G \left (\int _b ^y g\right )}{\partial \left (\int _b ^y g\right ) }\right ) \left (\frac{\partial }{\partial y}\int _b ^y g\right )
= g\left (\int _b ^y g\right )\frac{\partial }{\partial y}\left (G(y) - G(b) \right )
= g\left (\int _b ^y g\right )g(y)
f(x, y) = \int _a ^{\int _b ^y g} g.
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I got:
D_1f(x, y) = 0
D_2f(x, y) = \frac{\partial f}{\partial y}
= \frac{\partial }{\partial y} \int _a ^{\int _b ^y g}g
Let G be the antiderivative of g:
= \frac{\partial }{\partial y}\left ( G \left (\int _b ^y g\right ) - G(a) \right )
= \frac{\partial }{\partial y} G \left (\int _b ^y g\right )
= \left ( \frac{\partial G \left (\int _b ^y g\right )}{\partial \left (\int _b ^y g\right ) }\right ) \left (\frac{\partial }{\partial y}\int _b ^y g\right )
= g\left (\int _b ^y g\right )\frac{\partial }{\partial y}\left (G(y) - G(b) \right )
= g\left (\int _b ^y g\right )g(y)