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...
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.