Algebraic Differentiability Theorem

Let $f, g : I \to R$ where $I$ is an interval, $c \in I$ . Suppose $f, g$ are both differentiable at $c$ . Then $\forall α \in R$ then:

$α f$
$f + g$
$f g$
$\frac{f}{g}$ (assuming $g (c) \neq 0$ )
Are all differentiable at $c$ . Further we can say:
$(α f)^{'} (c) = α f^{'} (c)$
$(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$
$(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$
If $g (c) \neq 0$ then:

{(\frac{f}{g})}^{'} (c) = \frac{f^{'} (c) g (c) - f (c) g^{'} (c)}{g (c)^{2}}

Proof

\begin{aligned} (α f)^{'} (c) & = lim_{x \to c} \frac{(α f) (x) - (α f) (c)}{x - c} \\ = lim_{x \to c} \frac{α f (x) - α f (c)}{x - c} \\ = lim_{x \to c} (\frac{α f (x) - f (c)}{x - c}) \\ = α lim_{x \to c} \frac{f (x) - f (c)}{x - c} & ALT \\ = α f^{'} (c) \end{aligned}

\begin{aligned} (f + g)^{'} (c) & = lim_{x \to c} \frac{(f + g) (x) - (f + g) (c)}{x - c} \\ = lim_{x \to c} \frac{f (x) + g (x) - f (x) - g (x)}{x - c} \\ = lim_{x \to c} (\frac{f (x) - f (c)}{x - c} + \frac{g (x) - g (c)}{x - c}) \\ = lim_{x \to c} \frac{f (x) - f (c)}{x - c} + lim_{x \to c} \frac{g (x) - g (c)}{x - c} & ALT \\ = f^{'} (c) + g^{'} (c) \end{aligned}

Notice here that for the $lim_{x \to c} g (x) = g (c)$ because Differentiability Implies Continuity:

\begin{aligned} (f g)^{'} (c) & = lim_{x \to c} \frac{(f g) (x) - (f g) (c)}{x - c} \\ = lim_{x \to c} \frac{f (x) g (x) - f (c) g (c)}{x - c} \\ = lim_{x \to c} \frac{f (x) g (x) - f (c) g (x) + g (x) f (c) - g (c) f (c)}{x - c} \\ = lim_{x \to c} \frac{f (x) - f (c)}{x - c} \cdot g (x) + \frac{g (x) - g (c)}{x - c} \cdot f (c) \\ = lim_{x \to c} \frac{f (x) - f (c)}{x - c} \cdot g (x) + lim_{x \to c} \frac{g (x) - g (c)}{x - c} \cdot f (c) & ALT \\ = lim_{x \to c} (\frac{f (x) - f (c)}{x - c}) lim_{x \to c} g (x) + lim_{x \to c} (\frac{g (x) - g (c)}{x - c}) lim_{x \to c} f (c) & ALT \\ = f^{'} (c) g (c) + g^{'} (c) f (c) \end{aligned}

\begin{aligned} {(\frac{f}{g})}^{'} (c) & = lim_{x \to c} (\frac{(\frac{f}{g}) (x) - (\frac{f}{g} (c))}{x - c}) \\ = lim_{x \to c} \frac{\frac{f (x)}{g (x)} - \frac{f (c)}{g (c)}}{x - c} \\ = lim_{x \to c} \frac{f (x) g (c) - f (c) g (x)}{g (x) g (c) (x - c)} \\ = lim_{x \to c} \frac{f (x) g (c) - f (c) g (c) + f (c) g (c) - f (c) g (x)}{g (x) g (c) (x - c)} \\ = lim_{x \to c} \frac{1}{g (x) g (c)} \cdot (\frac{f (x) - f (c)}{x - c} \cdot g (c) - f (x) \frac{g (x) - g (c)}{x - c}) \\ = lim_{x \to c} \frac{1}{g (x) g (c)} \cdot (f^{'} (c) g (c) - f (c) g^{'} (c)) & ALT \\ = \frac{f^{'} (c) g (c) - f (c) g^{'} (c)}{g (c)^{2}} \end{aligned}

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