Please just check this [Analysis problem]

[AKG]

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)

 

[wais]

= g\left (\int _b ^y g\right )\frac{\partial }{\partial y} \left (\int _b ^y g\right )

= g\left (\int _b ^y g\right )g(y)

= g \left (\int _b ^y g\right )g(y)

Therefore, the first-order partial derivative with respect to y is g \left (\int _b ^y g\right )g(y).