Subsection 2.5.1 Proof of the Linearity of Differentiation (Theorem 2.4.2)
Recall that in Theorem 2.4.2 we defined \(S(x)=\alpha\,f(x)+\beta\,g(x)\text{,}\) where \(\alpha,\beta\in\bbbr\) are constants. We wish to compute \(S'(x)\text{,}\) so we start with the definition:
\begin{align*}
S'(x) &= \lim_{h \to 0} \frac{S(x+h)-S(x)}{h}
\end{align*}
Let us concentrate on the numerator of the expression inside the limit and then come back to the full limit in a moment. Substitute in the definition of \(S(x)\text{:}\)
\begin{align*}
S(x+h)-S(x) &=
\big[ \alpha f(x+h) + \beta g(x+h) \big] -
\big[ \alpha f(x) + \beta g(x) \big] &\text{collect terms}\\
&=\alpha\big[f(x+h)-f(x)]+\beta\big[g(x+h)-g(x)\big]
\end{align*}
Now it is easy to see the structures we need — namely, we almost have the expressions for the derivatives \(f'(x)\) and \(g'(x)\text{.}\) Indeed, all we need to do is divide by \(h\) and take the limit. So let's finish things off.
\begin{align*}
S'(x) &= \lim_{h \to 0} \frac{S(x+h)-S(x)}{h} & \text{from above}\\
&= \lim_{h \to 0} \frac{\alpha\big[f(x+h)-f(x)]+\beta\big[g(x+h)-g(x)\big] }{h}\\
&= \lim_{h \to 0} \left[ \alpha\frac{f(x+h)-f(x)}{h} + \beta\frac{g(x+h)-g(x)}{h}
\right] &\text{limit laws}\\
&= \alpha\lim_{h \to 0} \frac{f(x+h)-f(x)}{h} + \beta\lim_{h\to0}\frac{g(x+h)-g(x)}{h}\\
&=\alpha f'(x) + \beta g'(x)
\end{align*}
as required.