Home page for accesible maths Math 101 Chapter 3: Differentiation 3.6 Leibniz’s rules for derivatives 3.8 Proof of Leibniz’s product rule.

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3.7 Proof of Leibniz’s rules

To prove (iii) and (iv), we use the following.

LEMMA

A differentiable function is continuous.

Proof. To prove that $g$ is continuous at $a$ , we wish to prove that $g(a+h)\rightarrow g(a)$ as $h\rightarrow 0$ . For $g$ differentiable at $a$ , we introduce the difference quotient by

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

\rightarrow g^{\prime}(a)0+g(a)=g(a)

as $h\rightarrow 0$ ; hence $g$ is continuous at $a$ .