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The question I am trying to solve is the following:

Suppose $f(x)=x^2+(b+1)x+2c,\quad f(2)=4,$ and $f'(-1)=2.$ Find the constants $b$ and $c$.

I assume to solve this equation, I use a simultaneous equation of both $f(x)$ and $f '(x)$, therefor I'm trying to figure out how to find the derivative of $f (x)$.

I've learnt how to find $f '(x)$ of basic functions, where $x$ is the only variable.

However, I need to know: do I treat the unkown constants, $b$ and $c$, as I would $x$? For example, the derivative of $x^2$ is $2x$, would $2c$ become just $2$?

Note* I'm new to calculus

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    The derivative of a constant is zero. For example, $2c$ become $0$. You only treat $x$ like that, all others are constants. – 2017-01-28
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    in addition to previous comment, derivative of $(b+1)x$ is $(b+1)$ – 2017-01-28
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    Treat the constants as though they were fixed numbers. The derivative of 2x8 is 0. So the derivative of 2c is 0. The derivative is to measure the rate of change in f (x) in comparison of the rate of change is x. b and c are not changing at all. So the are fixed constants. – 2017-01-28

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The unknowns $b$ and $c$ are constants, i.e. they have a fixed value, you just don't know it yet. This differs from $x$ in that $x$ could be anything we want it to really, since $f(x)$ means $f$ is a function with which we allow $x$ to run wild and do what it wants. For your problem, you can take the derivative as if $b+1$ or $2c$ were the same thing as $\pi$ or $0$ or $5$, as they're all just constants.

$$f(x)=x^2+(b+1)x+2c$$ $$f'(x)=2x+b+1$$ $$f'(-1)=-2+b+1=b-1=2 \implies b=3$$ $$f(2)=2^2+(3+1)(2) + 2c = 4 \implies c=-4$$

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$f'(x)=2x+b+1$. By the conditions given we get $$4+2(b+1)+2c=4$$ and $$b-1=2.$$ Solving the both equations we get $b=3$, $c=-4$.