Use the definition of the derivative to prove that if \(f\) and \(g\) are differentiable at \(x\), then \(fg\) is differentiable at \(x\).
Solution
1. Write the difference quotient of \(fg\).
2. \(g\) is continuous at \(x\).
\(g\) is differentiable at \(x\), hence continuous at \(x\): \(g(x+h) \to g(x)\) as \(h \to 0\).
3. Pass to the limit.
as \(h \to 0\), \(\frac{f(x+h)-f(x)}{h} \to f'(x)\), \(g(x+h) \to g(x)\) (<1>2), and \(\frac{g(x+h)-g(x)}{h} \to g'(x)\); substituting into 1, \[ \lim_{h\to 0}\frac{(fg)(x+h)-(fg)(x)}{h} = f'(x)g(x) + f(x)g'(x), \] so \(fg\) is differentiable at \(x\) with derivative \(f'(x)g(x) + f(x)g'(x)\).
4. Q.E.D.