How to differentiate $ \exp \left\{ \int^t_sf(u)du \right\} $ w.r.t. $t$?

Question

I have looked and looked in my calculus books but couldn't find the answer to a problem I am facing of of the following form:

$$g(t) = \exp \left\{ \int^t_sf(u)du \right\} \\ \frac{\partial g(t)}{\partial t} = \ ? $$

How do I differentiate $g(t)$ w.r.t $t$?

$s$ is fixed.

Answer 1

It is a composite of the function $h(t):= \exp(t)$ and $k(t):= \int_s^t f(u) \, \mathrm{d} u$. By the chain rule we have $g'(t) = k'(t) h'(k(t))$. On the other hand, the fundamental theorem of calculus says that $k'(t) = f(t)$. Both together imply that $$g'(t) = k'(t) h'(k(t)) = f(t) \exp(k(t)) = f(t) g(t).$$ Note that this argument also proves that $g$ is differentiable. This steps are valid provided that $f$ is contiuous.

Answer 2

Chain rule; we know how to differentiate $\exp$ and we know how to differentiate $\int_s^t f(u)\,du$ (the Fundamental Theorem), so we just combine those with the chain rule..

Answer 3

We have $$\frac{\partial g(t)}{\partial t}{=\exp\left\{ \int^t_sf(u)du \right\}\cdot {\partial \int^t_sf(u)du\over \partial t}\\=g(t)\cdot \Big(f(t)-f(s)\Big)}$$can't proceed any further.