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3.8 Proof of Leibniz’s product rule.

(Not examinable)

(iii) In each case, we consider the difference quotient, and here we have

{{f(a+h)g(a+h)-f(a)g(a)}\over{h}}

=\Bigl({{f(a+h)-f(a)}\over{h}}\Bigr)g(a+h)+f(a)\Bigl({{g(a+h)-g(a)}\over{h}}\Bigr)

\rightarrow f^{\prime}(a)g(a)+f(a)g^{\prime}(a)\qquad(h\rightarrow 0)

where the limits exist since $f$ and $g$ are differentiable at $a$ and we have used the Lemma to deal with $g(a+h).$ Hence

(fg)^{\prime}(a)=f^{\prime}(a)g(a)+f(a)g^{\prime}(a).