Proof of Sum/Difference of Two Functions : (f(x) + g(x))′  = f ′(x) +  g ′(x)

 It is easy adequate to prove by using the definition of the derivative.  We will start with the sum of two functions. Firstly plug the sum in the definition of the derivative and rewrite the numerator a bit.

(f(x) + g(x))' = lim_h→0 (f(x + h) + g(x + h) - (f(x) + g(x)))/h

= lim_h→0 (f(x + h) - f(x) + g(x + h) ­- g(x))/h

Then, break up the fraction in two pieces and recall the limit of a sum is the total of the limits. By using this fact we consider that we end-up with the definition of the derivative for all of the two functions.

(f(x) + g(x))' = lim_h→0 (f(x + h) - f(x))/h + lim_h→0 (g(x + h) - g(x))/h

= f'(x) + g'(x)

The proof of the difference of two functions in nearly the same therefore we'll provide this here without any clarification.

(f(x) + g(x))' = lim_h→0 (f(x + h) - g(x + h) - (f(x) - g(x)))/h

= lim_h→0 (f(x + h) - f(x) - (g(x + h) ­- g(x))/h

= lim_h→0 ((f(x + h) - f(x))/h) - ((g(x + h) ­- g(x))/h)

= f'(x) - g'(x)