%% Derivative %%

Quotient Rule

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Cegep1

Mathematics

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214 words

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2 minutes

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

$$\frac{d}{dx}\frac{f(x)}{g(x)}=\frac{g(x)f^{\prime }(x)-f(x)g^{\prime }(x)}{(g(x){)}^2}\text{, provided }g(x)\ne 0$$

Proof

Let $q(x)=\frac{f(x)}{g(x)}$, then note that

$$\begin{array}{rl}q(x+h)-q(x)& =\frac{f(x+h)}{g(x+h)}-\frac{f(x)}{g(x)}=\frac{f(x+h)g(x)-f(x)g(x+h)}{g(x+h)g(x)}\\ ⟹\frac{q(x+h)-q(x)}{h}& =\frac{1}{h}\cdot \frac{f(x+h)g(x)-f(x)g(x+h)}{g(x+h)g(x)}\end{array}$$

Now,

$$\begin{array}{rl}\frac{\mathrm{d}}{\mathrm{d}x}\frac{f(x)}{g(x)}=q^{\prime }(x)& =\underset{h\to 0}{lim}\frac{q(x+h)-q(x)}{h}\\ & =\underset{h\to 0}{lim}\frac{1}{h}\cdot \frac{f(x+h)g(x)-f(x)g(x+h)}{g(x+h)g(x)}\\ ⟹& =\underset{h\to 0}{lim}\frac{1}{h}\cdot \frac{f(x+h)g(x)-f(x)g(x)+f(x)g(x)-f(x)g(x+h)}{g(x+h)g(x)}\\ & =\underset{h\to 0}{lim}\frac{1}{g(x+h)g(x)}\cdot \frac{f(x+h)g(x)-f(x)g(x)+f(x)g(x)-f(x)g(x+h)}{h}\\ & =\underset{h\to 0}{lim}\frac{1}{g(x+h)g(x)}(\frac{f(x+h)g(x)-f(x)g(x)}{h}+\frac{f(x)g(x)-f(x)g(x+h)}{h})\\ & =\underset{h\to 0}{lim}\frac{1}{g(x+h)g(x)}(g(x)\frac{f(x+h)-f(x)}{h}-f(x)\frac{g(x+h)-g(x)}{h})\\ & =\frac{1}{g(x)\underset{h\to 0}{lim}g(x+h)}(g(x)\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}-f(x)\underset{h\to 0}{lim}\frac{g(x+h)-g(x)}{h})\\ & =\frac{1}{g(x)g(x)}(g(x)f^{\prime }(x)-f(x)g^{\prime }(x))\\ & =\frac{g(x)f^{\prime }(x)-f(x)g^{\prime }(x)}{(g(x){)}^2}\end{array}$$

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Contributors

Ajitani Hifumi

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