I have a probably rather simple question, but I'm afraid I can't find the solution myself, even though it probably exists alredy somewhere in this board. I'm sorry to waste your time.
$\frac{d}{dy} \int_y^a \int_z^a f(y,z) \;dy\; dz$
where a is some finite number.
Thank you very much for your help!
Vils
On
As written, the interior $y$ is irrelevant because is a bound variable: $$ \frac{d}{dy}\int_y^a\int_z^a f(y,z)\,dy\,dz = \frac{d}{dy}\int_y^a\int_z^a f(x,z)\,dx\,dz = \frac{d}{dy}\int_a^y\int_a^z f(x,z)\,dx\,dz. $$ and now, applying the FTC: $$ \frac{d}{dy}\int_a^y\int_a^z f(x,z)\,dx\,dz = \int_a^y f(x,y)\,dx. $$
There is some abuse of notation here, insofar as the letter $y$ is used as integration variable in the inner integral and as independent variable for some "outer" function $h$. We are talking about $$h(y):=\int_y^a\int_z^a f(t,z)\>dt\>dz\ .$$ Let the inner integral be $$g(z):=\int_z^af(t,z)\>dt\ .$$ Then $$h(y)=\int_y^a g(z)\>dz$$ and therefore $$h'(y)=-g(y)=-\int_y^a f(t,y)\>dt\ .$$