Editing Linearity of differentiation (section)

===Sum===
Let <math>f, g</math> be functions. Let <math>j</math> be a function, where <math>j</math> is defined only where <math>f</math> and <math>g</math> are both defined.
(In other words, the domain of <math>j</math> is the intersection of the domains of <math>f</math> and <math>g</math>.) Let <math>x</math> be in the domain of <math>j</math>. Let <math>j(x) = f(x) + g(x)</math>.

We want to prove that <math>j^{\prime}(x) = f^{\prime}(x) + g^{\prime}(x)</math>.

By definition, we can see that

<math display="block">\begin{align}
j^{\prime}(x) &= \lim_{h \rightarrow 0} \frac{j(x + h) - j(x)}{h} \\
&= \lim_{h \rightarrow 0} \frac{\left( f(x + h) + g(x + h) \right) - \left( f(x) + g(x) \right)}{h} \\
&= \lim_{h \rightarrow 0} \left( \frac{f(x + h) - f(x)}{h} + \frac{g(x + h) - g(x)}{h} \right) \\
\end{align}</math>In order to use the law for the sum of limits here, we need to show that the individual limits, <math display="inline">\lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h}</math> and <math display="inline">\lim_{h \rightarrow 0} \frac{g(x + h) - g(x)}{h}</math> both exist. By definition, <math display="inline">f^{\prime}(x) = \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h}</math>and <math display="inline">g^{\prime}(x) = \lim_{h \rightarrow 0} \frac{g(x + h) - g(x)}{h}</math>, so the limits exist whenever the derivatives <math>f^{\prime}(x)</math> and <math>g^{\prime}(x)</math> exist. So, assuming that the derivatives exist, we can continue the above derivation

<math display="block">\begin{align}
j^{\prime}(x) &= \lim_{h \rightarrow 0} \left( \frac{f(x + h) - f(x)}{h} + \frac{g(x + h) - g(x)}{h} \right) \\
&= \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h} + \lim_{h \rightarrow 0} \frac{g(x + h) - g(x)}{h} \\
&= f^{\prime}(x) + g^{\prime}(x)
\end{align}</math>


Thus, we have shown what we wanted to show, that: <math>j^{\prime}(x) = f^{\prime}(x) + g^{\prime}(x)</math>.