Asked by Marissa
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.
Answers
Answered by
Steve
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)-f(a) + g(a+h)-g(a))/h
= (f(a+h)+g(a+h) - (f(a)+g(a))/h
= ((f+g)(a+h) - (f+g)(a))/h
the limit is d(f+g)/dx at x=a
(f(a+h)-f(a))/h
(g(a+h)-g(a))/h
sum of limits is thus
(f(a+h)-f(a) + g(a+h)-g(a))/h
= (f(a+h)+g(a+h) - (f(a)+g(a))/h
= ((f+g)(a+h) - (f+g)(a))/h
the limit is d(f+g)/dx at x=a
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