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:

$$= \frac{\partial }{\partial y}\left ( G \left (\int _b ^y g\right ) - G(a) \right )$$

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

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

$$= g\left (\int _b ^y g\right )\frac{\partial }{\partial y}\left (G(y) - G(b) \right )$$

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

 

[anathema]

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

Therefore, the partial derivative with respect to y is g(y)g\left (\int _b ^y g\right ). The partial derivative with respect to x is 0 because the function does not depend on x.

 

[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).