(related to Proposition: Basis Arithmetic Operations Involving Differentiable Functions, Product Rule, Quotient Rule)
Let \(D\subseteq\mathbb R\) (\(D\) is a subset of real numbers). Let \(x\in D,\lambda\in\mathbb R\). By hypothesis, \(f,g:D\to\mathbb R\) are differentiable functions at \(x\).
This rule follows immediately the definition of derivatives and the calculation rule for the sum of convergent real sequences.
This rule follows immediately the definition of derivatives and the calculation rule for the difference of convergent real sequences.
This rule follows immediately the definition of derivatives and the calculation rule for the product of a real number with a convergent real sequence.
Applying the definition of derivatives we get
\[\begin{array}{rcl} (fg)'(x)&=&\lim_{h\to 0}\frac{f(x+h)g(x+h)-f(x)g(x)}{h}\\ &=&\lim_{h\to 0}\frac{f(x+h)g(x+h)-f(x+h)g(x)+f(x+h)g(x)-f(x)g(x)}{h}\\ &=&\lim_{h\to 0}\frac{f(x+h)(g(x+h)-g(x))+(f(x+h)-f(x))g(x)}{h}\\ &=&\lim_{h\to 0}f(x+h)\frac{g(x+h)-g(x)}h+\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}g(x)\quad\quad( * )\\ &=&f'(x)g(x) + f(x)g'(x). \end{array}\]
In the step \( ( * ) \) we have used that \(f\) is continuous, because it is differentiable at \(x\).
By hypothesis, we have \(g(\xi)\neq 0\) for all \(\xi\in D\). We first consider the special case \(f=1\).
\[\begin{array}{rcl} \left(\frac 1g\right)'(x)&=&\lim_{h\to 0}\frac 1h\left(\frac 1{g(x+h)}- \frac 1{g(x)}\right)\\ &=&\lim_{h\to 0}\frac {1}{g(x+h)g(x)}\left(\frac {g(x)-g(x+h)}h\right)\\ &=&\frac {-g'(x)}{g(x)^2}. \end{array}\] Now, we can derive the general case by applying the Product Rule \((4)\) above:
\[\begin{array}{rcl} \left(\frac fg\right)'(x)&=&\left(f\cdot \frac 1g\right)'(x)\\ &=&f'(x)\frac 1{g(x)} + f(x)\frac{-g'(x)}{g(x)^2}\\ &=&\frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}. \end{array}\]
