Editing Linearity of differentiation (section)

===Constant coefficient===
Let <math>f</math> be a function. Let <math>a \in \mathbb{R}</math>; <math>a</math> will be the constant coefficient. Let <math>j</math> be a function, where j is defined only where <math>f</math> is defined. (In other words, the domain of <math>j</math> is equal to the domain of <math>f</math>.) Let <math>x</math> be in the domain of <math>j</math>. Let <math>j(x) = af(x)</math>.

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

Now, in order to use a limit law for constant coefficients to show that

<math display="block">
\lim_{h \rightarrow 0} a\frac{f(x + h) - f(x)}{h} = a\lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h} 
</math> we need to show that <math display="inline">\lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h}</math> exists.
However, <math display="inline">f^{\prime}(x) = \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h}</math>, by the definition of the derivative. So, if <math>f^{\prime}(x)</math> exists, then  <math display="inline">\lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h}</math> exists.

Thus, if we assume that <math>f^{\prime}(x)</math> exists, we can use the limit law and continue our proof.

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

Thus, we have proven that when <math>j(x) = af(x)</math>, we have <math>j^{\prime}(x) = af^{\prime}(x)</math>.