Editing Linearity of differentiation (section)

===Linearity (directly)===
Let <math>a, b \in \mathbb{R}</math>. 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) = af(x) + bg(x)</math>.

We want to prove that <math>j^{\prime}(x) = af^{\prime}(x) + bg^{\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( af(x + h) + bg(x + h) \right) - \left( af(x) + bg(x) \right)}{h} \\
&= \lim_{h \rightarrow 0} \left( a\frac{f(x + h) - f(x)}{h} + b\frac{g(x + h) - g(x)}{h} \right) \\
\end{align}</math>


In order to use the limits law for the sum of limits, we need to know that <math display="inline">\lim_{h \to 0} a\frac{f(x + h) - f(x)}{h}</math> and <math display="inline">\lim_{h \to 0} b\frac{g(x + h) - g(x)}{h}</math> both individually exist. For these smaller limits, we need to know that <math display="inline">\lim_{h \to 0} \frac{f(x + h) - f(x)}{h}</math> and <math display="inline">\lim_{h \to 0} \frac{g(x + h) - g(x)}{h}</math> both individually exist to use the coefficient law for limits. By definition, <math display="inline">f^{\prime}(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}</math> and <math display="inline">g^{\prime}(x) = \lim_{h \to 0} \frac{g(x + h) - g(x)}{h}</math>. So, if we know that <math>f^{\prime}(x)</math> and <math>g^{\prime}(x)</math> both exist, we will know that <math display="inline">\lim_{h \to 0} \frac{f(x + h) - f(x)}{h}</math> and <math display="inline">\lim_{h \to 0} \frac{g(x + h) - g(x)}{h}</math> both individually exist. This allows us to use the coefficient law for limits to write

<math display="block">
\lim_{h \to 0} a\frac{f(x + h) - f(x)}{h}
= a\lim_{h \to 0}\frac{f(x + h) - f(x)}{h}
</math>

and

<math display="block">
\lim_{h \to 0} b\frac{g(x + h) - g(x)}{h}
= b\lim_{h \to 0}\frac{g(x + h) - g(x)}{h}.
</math>

With this, we can go back to apply the limit law for the sum of limits, since we know that <math display="inline">\lim_{h \rightarrow 0} a\frac{f(x + h) - f(x)}{h}</math> and <math display="inline">\lim_{h \rightarrow 0} b\frac{g(x + h) - g(x)}{h}</math> both individually exist. From here, we can directly go back to the derivative we were working on.<math display="block">\begin{align}
j^{\prime}(x) &= \lim_{h \rightarrow 0} \left( a\frac{f(x + h) - f(x)}{h} + b\frac{g(x + h) - g(x)}{h} \right) \\
&= \lim_{h \rightarrow 0} \left( a\frac{f(x + h) - f(x)}{h}\right) + \lim_{h \rightarrow 0} \left(b\frac{g(x + h) - g(x)}{h} \right) \\
&= a\lim_{h \rightarrow 0} \left( \frac{f(x + h) - f(x)}{h}\right) + b\lim_{h \rightarrow 0} \left(\frac{g(x + h) - g(x)}{h} \right) \\
&= af^{\prime}(x) + bg^{\prime}(x)
\end{align}</math>Finally, we have shown what we claimed in the beginning: <math>j^{\prime}(x) = af^{\prime}(x) + bg^{\prime}(x)</math>.