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As usual when differentiating something with respect to a variable that appears twice. The chain rule for partial derivatives.

For example, consider function $z = f(u,v)$. Suppose we want $(d/dt)f(t,t)$. Let $u=v=t$ and use $dz/dt = (\partial z/\partial u)(du/dt) + (\partial z/\partial v)(dv/dt)$.

Thus...Thus… $$ \frac{d}{dt}\int_0^t\int_0^t f(x,y)\\,dx\\,dy = \int_0^t f(t,y)\\,dy + \int_0^t f(x,t)\\,dx $$$$ \frac{d}{dt}\int_0^t\int_0^t f(x,y)\,dx\,dy = \int_0^t f(t,y)\,dy + \int_0^t f(x,t)\,dx $$

By the way, why did you write $\partial/\partial t$ to differentiate a function of the single variable $t$? It's not wrong, just confusing to students.

As usual when differentiating something with respect to a variable that appears twice. The chain rule for partial derivatives.

For example, consider function $z = f(u,v)$. Suppose we want $(d/dt)f(t,t)$. Let $u=v=t$ and use $dz/dt = (\partial z/\partial u)(du/dt) + (\partial z/\partial v)(dv/dt)$.

Thus... $$ \frac{d}{dt}\int_0^t\int_0^t f(x,y)\\,dx\\,dy = \int_0^t f(t,y)\\,dy + \int_0^t f(x,t)\\,dx $$

By the way, why did you write $\partial/\partial t$ to differentiate a function of the single variable $t$? It's not wrong, just confusing to students.

As usual when differentiating something with respect to a variable that appears twice. The chain rule for partial derivatives.

For example, consider function $z = f(u,v)$. Suppose we want $(d/dt)f(t,t)$. Let $u=v=t$ and use $dz/dt = (\partial z/\partial u)(du/dt) + (\partial z/\partial v)(dv/dt)$.

Thus… $$ \frac{d}{dt}\int_0^t\int_0^t f(x,y)\,dx\,dy = \int_0^t f(t,y)\,dy + \int_0^t f(x,t)\,dx $$

By the way, why did you write $\partial/\partial t$ to differentiate a function of the single variable $t$? It's not wrong, just confusing to students.

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Gerald Edgar
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As usual when differentiating something with respect to a variable that appears twice. The chain rule for partial derivatives.

For example, consider function $z = f(u,v)$. Suppose we want $(d/dt)f(t,t)$. Let $u=v=t$ and use $dz/dt = (\partial z/\partial u)(du/dt) + (\partial z/\partial v)(dv/dt)$.

Thus... $$ \frac{d}{dt}\int_0^t\int_0^t f(x,y)\\,dx\\,dy = \int_0^t f(t,y)\\,dy + \int_0^t f(x,t)\\,dx $$

By the way, why did you write $\partial/\partial t$ to differentiate a function of the single variable $t$? It's not wrong, just confusing to students.