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I need to evaluate this expression:

$$\frac{d}{dt}\exp\left({-\int_t^T r(\tau) d\tau}\right)$$

I first start with the exponential rule

$$\frac{d}{dx}e^{g(x)} = \frac{dg(x)}{dx} e^{g(x)}$$

Now I need to evaluate

$$-\frac{d}{dt}\int_t^T r(\tau) d\tau$$

My first thought was to use Leibniz integral rule, but I can't use the simple form

$$\frac{d}{dt}\int_a^b f(x,t)dx=\int_a^b \frac{\partial}{\partial t}f(x,t)dx$$

Because one of the limits is what I am differentiating by. So in fact it is the more general form that the problem reverts to and that complicates the result a lot more than I am expecting. What am I missing?

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    $\begingroup$ $\frac d{dx}\int_c^x f(u)\;du = f(x)$. Note your variable of interest only appears in the limits of integration, so you don't need Leibniz. Note also you will have to switch the limits (and introduce a minus sign) to use this fact. $\endgroup$ – MPW Oct 3 '18 at 14:25
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Thanks @MPW I think I got it

$$-\frac{d}{dt}\int_t^T r(\tau) d\tau$$ $$=\frac{d}{dt}\int_T^t r(\tau) d\tau = r(t)$$

Putting this back into the original problem I get

$$r(t)\exp\left(-\int_t^T r(\tau) d\tau\right)$$

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Assuming $T$ is constant with respect to $t$:

$- \frac{d}{dt} \int_{t}^{T} r(\tau) d\tau = \frac{d}{dt} \int_{T}^{t} r(\tau)d\tau =\frac{d}{dt}(R(t))-\frac{d}{dt}(R(T))$.

Because $R(T)$ is constant with respect to $t$:

$\frac{d}{dt}(R(T))=0$.

Therefore: $\frac{d}{dt}(R(t))-\frac{d}{dt}(R(T))=\frac{d}{dt}R(t)=r(t)$

All together:

$- \frac{d}{dt} \int_{t}^{T} r(\tau) d\tau=r(t)$

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