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Product Rule

Tags

Calculus

Cegep/1

Word count

170 words

Reading time

2 minutes

Let $f$ and $g$ be two differentiable functions, then

$$\frac{d}{dx}(f(x)g(x))=f(x)g^{\prime }(x)+g(x)f^{\prime }(x)$$

The rule extends to >2 functions:

$$\frac{d}{dx}(f(x)g(x)h(x))=f^{\prime }(x)g(x)h(x)+g^{\prime }(x)f(x)h(x)+h^{\prime }(x)f(x)g(x)$$

Proof

Let $p(x)=f(x)g(x)$, then

$$\begin{array}{rl}\frac{\mathrm{d}}{\mathrm{d}x}(f(x)g(x))=p^{\prime }(x)& =\underset{h\to 0}{lim}\frac{p(x+h)-p(x)}{h}\\ & =\underset{h\to 0}{lim}\frac{f(x+h)g(x+h)-f(x)g(x)}{h}\\ ⟹& =\underset{h\to 0}{lim}\frac{f(x+h)g(x+h)-f(x+h)g(x)+f(x+h)g(x)-f(x)g(x)}{h}\\ & =\underset{h\to 0}{lim}\frac{f(x+h)(g(x+h)-g(x))+g(x)(f(x+h)-f(x))}{h}\\ & =\underset{h\to 0}{lim}(\frac{f(x+h)(g(x+h)-g(x))}{h}+\frac{g(x)(f(x+h)-f(x))}{h})\\ & =\underset{h\to 0}{lim}\frac{f(x+h)(g(x+h)-g(x))}{h}+\underset{h\to 0}{lim}\frac{g(x)(f(x+h)-f(x))}{h}\\ & =\underset{h\to 0}{lim}f(x+h)\underset{h\to 0}{lim}\frac{g(x+h)-g(x)}{h}+g(x)\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}\\ & =f(x)g^{\prime }(x)+g(x)f^{\prime }(x)\end{array}$$

Contributors

Ajitani Hifumi

Changelog

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