Can someone explain to me the difference between the results of $ A$ and $B$, where $$A=\frac{d}{dc} \int_{-\infty}^c xf(x) dx $$ $$B= \frac{d}{dc} \int_c^{+\infty} xf(x) dx $$
You can image $f(x)$ to be the probability density function of X random variable. So that $F(c) = \int_{-\infty}^c f(x)$
Is it correct to write A as follows $$ A=\frac{d}{dc}\Big[x \int_{-\infty}^c f(x)dx -\int_{-\infty}^c \int_{-\infty}^c f(x)dx \Big]\\ =\frac{d}{dc}x \int_{-\infty}^c f(x) dx -\frac{d}{dc}\int_{-\infty}^c \int_{-\infty}^c f(x)dx\\ =xf(c)-F(c) $$
and B as follows $$ B= \frac{d}{dc}\Big[x \int_c^{+\infty} f(x)dx -\int_c^{+\infty} \int_c^{+\infty} f(x)dx \Big]\\ =\frac{d}{dc}x \int_c^{+\infty} f(x) dx -\frac{d}{dc}\int_c^{+\infty} \int_c^{+\infty} f(x)dx\\ =\frac{d}{dc}x [1-F(c)] - [1-F(c)]\\ =0+xf(c)-1+F(c) $$
Your derivation makes little sense. There is no way for the variable $x$ to leave the integrals like that and those double integrals are not syntactically correct.
Under the assumption that $f$ is a continuous pdf for a random variable $X$ with finite expectation (hence the integrals converge) you can directly apply the Fundamental Theorem of Calculus (which does apply to convergent improper integrals, see this question) to get that:
$$A = cf(c)$$
and
$$B = -cf(c).$$