It does not a priori make sense to differentiate x! because the domain of x\mapsto x! is \mathbf N, not \mathbf R (or anything else supporting a good notion of differentiation, like \mathbf C ...

Use the chain rule. Define u = x + c then use the fact that \frac{d\cdot}{dx} = \frac{du}{dx} \frac{d\cdot}{du} where the \cdot represents any function, so \frac{df}{dx} = \frac{du}{dx} \frac{df}{du} ...

The symbols d/dx and x should both be interpreted as linear operators acting on a vector space that the unknown function y belongs to. The sum of linear operators is well-defined and that is ...

Not sure about the problem but the strength of the electrical field, E, depends on your distance from it, which I assume is x. \frac{dE}{dx} then, is how much the strength of the field changes ...

Everything is correct, except that the derivative of a constant (like 6) is always 0. You can still see this fact from the power rule. Write 6 as 6x^0. The power rule says that the derivative is 6 \cdot 0 x^{-1} ...

If you choose u=x^2+1, then taking the derivative with respect to x gives: \frac{\textrm{d}u}{\textrm{d}x}=2x. Therefore \textrm{d}u=2x\textrm{d}x or x\textrm{d}x=\textrm{d}u/2. Now your ...