Assuming that f and g are functions differentiable at a (though we do not know their formulas). Prove that f +g is differentiable at a using the definition of the derivative.
we know these limits exist as h->0:
(f(a+h)-f(a))/h
(g(a+h)-g(a))/h
sum of limits is thus
(f(a+h)-(a+h)-g(a))/h
= (f(a+h)+g(a+h) - ((a))/h
= ((f+g)(a+h) - (f+g)(a))/h
the limit is d(f+g)/dx at x=a
To prove that the sum of two differentiable functions, f and g, is differentiable at a, we need to show that the limit of the difference quotient exists as x approaches a.
Let's start by considering the difference quotient for the function f+g at a:
[f(x) + g(x) - ((a))] / (x - a)
This can be simplified as:
[f(x) - f(a)] / (x - a) + [g(x) - g(a)] / (x - a)
Now, since f and g are differentiable at a, we can use the definition of the derivative to express the difference quotients in terms of the derivatives:
[f(x) - f(a)] / (x - a) = f'(a) + ε1(x)
[g(x) - g(a)] / (x - a) = g'(a) + ε2(x)
Here, ε1(x) and ε2(x) represent the error terms which go to 0 as x approaches a.
Substituting these expressions back into our original difference quotient:
[f(x) + g(x) - ((a))] / (x - a) = (f'(a) + ε1(x)) + (g'(a) + ε2(x))
By using algebraic properties, we get:
[f(x) + g(x) - ((a))] / (x - a) = '(a) + ε1(x) + ε2(x)
Now, let's define ε(x) = ε1(x) + ε2(x). Since ε1(x) and ε2(x) go to 0 as x approaches a, ε(x) also goes to 0 as x approaches a.
Hence, we have:
[f(x) + g(x) - ((a))] / (x - a) = '(a) + ε(x)
Finally, we can see that as x approaches a, the difference quotient approaches '(a), which means that the derivative of f + g exists at a. Thus, we have proven that f + g is differentiable at a using the definition of the derivative.