Well I am having a bad time with this derivative, because I cannot understand how to proceed when you have two functions inside the integral with two different variables.
$$f(x)= \int_{\sin(\cos(x))}^{{\textstyle\int_1^2} t\sin(\cos(t)) \,\mathrm dt} e^xh(t) \,\mathrm dt$$ And if $h(x)$ is a continuous function, then find $\dfrac{\mathrm df(x)}{\mathrm dx}$.
Thanks to everyone.
On
Since you oddly don't even know what $h$ is, other than hopefully being continuous, then you are only really interested in finding the derivative for $f$ in terms of $h$, rather than the function $f$ itself.
Duely note that the definite integral $\int_1^2 t\sin\cos t~\mathrm d t$ is a constant (assuming that it converges). It is just a distraction.
Now, for any constant $c$, differentiable function $g$, and continuous function $h$, the fundamental principles of calculus inform that:
$$\dfrac{\mathrm d ~~}{\mathrm d x}\int_{g(x)}^c h(t)~\mathrm d t = {-h(g(x))\dfrac{\mathrm d g(x)}{\mathrm d x\quad}}$$
And so you can now find $\displaystyle\dfrac{\mathrm d ~~}{\mathrm d x}~ e^x\int_{g(x)}^c h(t)~\mathrm d t$.
Things will work out nicely if you pretend that you know an anti-derivative $H(x)$ for $h(x).$ Since you're integrating with respect to $t$, you can pull the $e^x$ outside. Then you get:
$$f(x) = e^x \int_{\sin(\cos x)}^{\int_1^2 \sin(\cos t)) \; dt} h(t) \; dt = e^x\left( H\left(\int_1^2 \sin(\cos t)) \; dt\right) - H\left(\sin(\cos x)\right)\right)=e^x\mbox{junk}$$
The first argument for $H$ is a constant, so it's easy to take the derivative of $f$ now:
$$\frac{df}{dx} = e^x\mbox{junk} + e^x\frac{d\; \mbox{junk}}{dx} = f(x) +e^x\frac{d\; \mbox{junk}}{dx}.$$
The derivative of "junk" is straightforward. The first term is constant and the second term is two applications of chain rule:
$$\frac{d\; \mbox{junk}}{dx} = 0 - H'(\sin(\cos x))\cos(\cos(x))(-\sin(x)).$$
And since $H' = h$, we can put it all together to write:
$$\frac{df}{dx} = f(x) +e^x\left( - h(\sin(\cos x))\cos(\cos(x))(-\sin(x)) \right).$$