TITLE: Derivative of an integral with three variables
QUESTION [1 upvotes]: I am having problems trying to make the correct composition of functions to apply the chain rule. Any advice?
Here is the problem

Let:
  $$f(x)=sin\left(\int_{\sin(\cos(x))}^{{\textstyle\int_1^2} t\sin(\cos(t)) \,\mathrm dt} \left(\int_2^{\sin(u)} e^{\sin(t)} \,\mathrm dt\right) \,\mathrm du\right)$$
  Then find $\dfrac{\mathrm df(x)}{\mathrm dx}$

My try was to propose the functions:
$$F(u) =\int_{sin(cos(x))}^{{\textstyle\int_1^2} t\sin(\cos(t)) \,\mathrm dt} u \,\mathrm du$$
And
$$G(u) =\int_2^{\sin(u)} e^{\sin(t)} \,\mathrm dt$$
So $f(x) = F\left(G(u)\right)$ But that is not function of x...

REPLY [1 votes]: Let's make some substitutions to make this more digestible:
$$ a(x) = \sin(\cos x) \\b = \int_1^2sa(s)\mathrm ds \approx. 0.01881 \text{ (constant)} \\ h(u) = \int_2^{\sin u} e^{\sin t}\mathrm dt \\ g(x) = \int_{a(x)}^b h(u)\mathrm du \\ f(x) = \sin(g(x))  $$
Then we can apply the chain rule repeatedly with FTC:
$$ f'(x) = \cos(g(x))g'(x) \\ g'(x) = -h(a(x))a'(x) \\ a'(x) = -\cos(\cos x)\sin x$$
Substituting back in the derivatives:
$$ f(x) = \cos(g(x))h(a(x))\cos(\cos x)\sin x $$
and finally substituting in all the values:
$$ f(x) = \cos\left(\int_{\sin(\cos x)}^{\int_1^2s\sin(\cos s)\mathrm ds} \left(\int_2^{\sin u} e^{\sin t}\mathrm dt\right) \mathrm du \right)\left(\int_2^{\sin(\sin(\cos x))} e^{\sin t}\mathrm dt\right) \cos(\cos x)\sin x  $$

Edit: I changed the dummy variable in the $b$ integral from $t$ to $s$ so that it's clearly a different variable from the dummy variable in the $h$ integral.
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