Proof of Differentiation Rules - The Derivative of the Sum of Two Functions: Calculus 1 Project by Nick Woodward
Overview
This Project has been completed as part of a standard 10 weeks Calculus 1 asynchronous online course with optional WebEx sessions during Summer 2021 Semester at MassBay Community College, Wellesley Hills, MA.
Proof of Differentiation Rules - The Derivative of the Sum of Two Functions
The Derivative of the Sum of Two Functions
\([f(x)+g(x)]'=f'(x)+g'(x)\)
To start, we first need to review the definition of a derivative:
$$ f'(x)= \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$$
Next, we set up our equation that we will solve to prove the rule of the derivative of the sum of 2 functions
$$ [f(x)+g(x)]'= \lim_{h \to 0} \dfrac{(f(x+h)+g(x+h))-(f(x)+g(x))}{h}$$
Once we have our variable substitued into our equation, we can distribute the negative value in the numerator
$$~~~~~~~~~~~~~~~~=\lim_{h \to 0} \dfrac{f(x+h)+g(x+h)-f(x)-g(x)}{h}$$
Next, we can arrange our variables in the numerator to be a sum of 2 fractions which leaves us with the following
$$~~~~~~~~~~~~~~~~~~~~~~=\lim_{h \to 0}\Big( \dfrac{f(x+h)-f(x)}{h}+\dfrac{g(x+h)-g(x)}{h}\Big)$$
After this, we can write this equation of the sum of 2 limits which yields the following equation
$$~~~~~~~~~~~~~~~~~~~~~~=\lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}+\lim_{h \to 0}\dfrac{g(x+h)-g(x)}{h}$$
Next, we can recognize that the following limits are the definition of the derivative for f'(x) and g'(x) respectively
$$~~~~~~~~~~~~~~~\lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}+\lim_{h \to 0}\dfrac{g(x+h)-g(x)}{h}=f'(x)+g'(x)$$
For further understanding of this topic, I highly recommend the following content:
The first is a video explaining the definition of a derivative
The next video explains the basic math behind this calculation and how we can separate the equation to be a sum of 2 fractions
I would also recommend this in depth page that explains derivative notation which is a fundemental concept of this material