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I am trying to show this using Leibnitz rule:

$$D_2f(x,y) = \frac{\partial {}}{\partial{y}} \left ( \int_0^xg_1 (t,0) \ dt + \int_0^y g_2(x,s) \ ds \right)$$ $$= \int_0^x \frac{\partial{}}{\partial{y}} g_1(t,0) \ dt + \int_0^y \frac{\partial{}}{\partial{y}} g_2(x,s) \ ds$$ $$ = \int_0^x \frac{\partial{}}{\partial{y}} g_1(t,0) \ dt + g_2(x,y) - g(x,0)$$

How do I calculate the first integral?

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  • $\begingroup$ Why not just use the FTC? $\endgroup$ – zhw. Nov 15 '15 at 17:50
  • $\begingroup$ @zhw. I am not sure how. I used it for the second integral. $\endgroup$ – realanalysis Nov 15 '15 at 17:50
  • $\begingroup$ When $x$ is fixed, you have something of the form $a(y) = C +\int_0^y b(s)\,ds.$ $\endgroup$ – zhw. Nov 15 '15 at 17:52
  • $\begingroup$ @zhw. I really don't follow what you're referencing too - which integral do you mean? $\endgroup$ – realanalysis Nov 15 '15 at 17:55
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Hint $$ \frac{\partial {}}{\partial{y}} \left ( \int_0^xg_1 (t,0) \ dt + \int_0^y g_2(x,s) \ ds \right)=0+g_2(x,y)=g_2(x,y)$$ because $\int_0^xg_1 (t,0) \ dt$ doesn't have any $y$, i.e. constant with respect to $y$.

also note that $$ \frac{\partial {}}{\partial{y}}\int_0^y g_2(x,s) \ ds\ne \int_0^y \frac{\partial {}}{\partial{y}}g_2(x,s) $$ but $$ \frac{\partial {}}{\partial{y}}\int_0^y g_2(x,s) \ ds = \frac{\partial {}}{\partial{y}}\left(G_2(x,y)-G_2(x,0)\right)=g_2(x,y) $$ where $G_2$ is antiderivative of $g_2$ with respect to $y$, i.e. $\frac{d}{dy}G_2(x,y)=g_2(x,y)$

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  • $\begingroup$ That's originally what I had also - but that implies $D_2f(x,y) = g_2(x,y) - g_2(x,0)$ which is incorrect unless $g_2(x,0) = 0$ $\endgroup$ – realanalysis Nov 15 '15 at 18:09
  • $\begingroup$ sorry, fixed it $\endgroup$ – Michael Medvinsky Nov 15 '15 at 18:32
  • $\begingroup$ could you explain the second part? why $d/dy G_2(x,0) = 0$? $\endgroup$ – FACEIT Nov 15 '15 at 18:33
  • $\begingroup$ @MichaelMedvinsky thanks I got it now - if you have time could you take a look at my second question I posted a while back? $\endgroup$ – realanalysis Nov 15 '15 at 18:35

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