Question: > 
> Find a number and a function that satisfies:
> $$\displaystyle\int\_{c}^{x}t\cdot f(t)\,\mathrm{d}x=x\sin (x)+\cos (x)+\dfrac{x^3}{3}$$
> 
> 
> 

I tried taking derivative both sides so I got:

$$\left(\displaystyle\int\_{0}^{x}t\cdot f(t)\,\mathrm{d}t-\displaystyle\int\_{0}^{c}t \cdot f(t)\,\mathrm{d}t\right)^\prime=x \cos(x)+x^2 $$

It's clear that the right part of the integral will be $0$ because it's constant:

$$\left( \displaystyle\int\_{0}^{x} t \cdot f(t)\,\mathrm{d}t \right)^{\prime}=x \cos(x)+x^2$$

But I didn't do rest. Because I don't know how to derive the integral with two functions. I know with one function like $\int\_{0}^{x} f(t)\,\mathrm{d}t$. Thanks!

Answer: 
First of all, when involving two variables, you should clarify the variable which you are differentiating. Following from your question, because of the fundamental theorem of calculus, we have:
$$
\frac{d\left( \displaystyle\int\_{0}^{x} t \cdot f(t)\mathrm{d}t \right)}{dx}=x \cos(x)+x^2
$$
Implies:
$$
xf(x)=x\cos(x)+x^2 \implies f(x)=\cos(x)+x
$$
Let's go back to the beginning. If we substitute the value of $f$, we get:
Implies:

$$
\left( \displaystyle\int\_{c}^{x} t \cdot (\cos(t)+t)\mathrm{d}t \right)=x\sin (x)+\cos (x)+\dfrac{x^3}{3}
$$
After doing all the calculations, we will have:
$$
c\sin (c)+\cos (c)+\dfrac{c^3}{3}=0
$$
I used an external program to solve this, but you can use Bolzano's Theorem to see that the equation has a solution for some $c\in [-1,0]$
You can still think of $t\mapsto tf(t)$ as a function, and differentiate as you normally would using the [fundamental theorem of calculus](https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus):
$$\frac{d}{dx}\left( \int\_{0}^{x} t \cdot f(t)\,\mathrm{d}t \right) = xf(x).$$