# Using Leibnitz Integral rule

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?

• Why not just use the FTC? – zhw. Nov 15 '15 at 17:50
• @zhw. I am not sure how. I used it for the second integral. – realanalysis Nov 15 '15 at 17:50
• When $x$ is fixed, you have something of the form $a(y) = C +\int_0^y b(s)\,ds.$ – zhw. Nov 15 '15 at 17:52
• @zhw. I really don't follow what you're referencing too - which integral do you mean? – realanalysis Nov 15 '15 at 17:55

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. $\frac{d}{dy}G_2(x,y)=g_2(x,y)$
• 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 $g_2(x,0) = 0$ – realanalysis Nov 15 '15 at 18:09
• could you explain the second part? why $d/dy G_2(x,0) = 0$? – FACEIT Nov 15 '15 at 18:33