Theorem: If   f   and   g   are differentiable then   d⁄dx(f(x)g(x)) = d⁄dx(f(x))g(x) + d⁄dx(g(x))f(x)  
Proof: We apply the definition of   d⁄dx, but then we need a trick, and we also use the result that if a function is differentiable then it is continuous.

  1.Insert here    = limh → 0 1⁄h(f(x + h)g(x + h) - f(x)g(x))  

  2.Insert here    = limh → 0(1⁄h(f(x + h) - f(x))g(x + h) + (g(x + h) - g(x))f(x))  

  3.Insert here    = limh → 0(1⁄h(f(x + h) - f(x))g(x + h) + 1⁄h(g(x + h) - g(x))f(x))  

  4.Insert here    = d⁄dx(f(x))g(x) + d⁄dx(g(x))f(x).

  5.Insert here    = limh → 0 1⁄h(f(x + h) - f(x))g(x + h) + limh → 0 1⁄h(g(x + h) - g(x))f(x)  

  6.Insert here   d⁄dx(F(x)) = limh → 0 1⁄h(F(x + h) - F(x))  

  7.Insert here    = limh → 0 1⁄h(f(x + h)g(x + h) - f(x)g(x + h) + f(x)g(x + h) - f(x)g(x))   (the trick!)

  8.Insert here Let   F(x) = f(x)g(x). Then

  9.Insert here    = limh → 0 1⁄h(f(x + h) - f(x))limh → 0 g(x + h) + limh → 0 1⁄h(g(x + h) - g(x))limh → 0 f(x)  

In the last step we used the fact that   limh → 0 g(x + h) = g(limh → 0(x + h)) = g(x)   since   g   is differentiable and hence continuous. Also   limh → 0 f(x) = f(x)   since   f(x)   is constant with respect to   h.

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