The Product and Quotient Rules

Product Rule: Let $u=f(x)$ and $v=g(x)$ be differentiable functions. Then $$(fg)’(x)=f(x)g’(x)+f’(x)g(x)$$ or $$\frac{d(uv)}{dx}=u\frac{dv}{dx}+\frac{du}{dx}v.$$

Proof. \begin{eqnarray*}(fg)’(x)&=&\lim_{\Delta x\to 0}\frac{fg(x+\Delta x)-fg(x)}{\Delta x}\\&=&\lim_{\Delta x\to 0}\frac{f(x+\Delta x)g(x+\Delta x)-f(x)g(x)}{\Delta x}\\&=&\lim_{\Delta x\to 0}\frac{f(x+\Delta x)g(x+\Delta x)-f(x)g(x+\Delta x)+f(x)g(x+\Delta x)-f(x)g(x)}{\Delta x}\\&=&\lim_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}g(x+\Delta x)+f(x)\lim_{\Delta x\to 0}\frac{g(x+\Delta x)-g(x)}{\Delta x}\\&=&f’(x)g(x)+f(x)g’(x).\end{eqnarray*} Note that $\displaystyle\lim_{\Delta x\to 0}g(x+\Delta x)=g(x)$ because $g(x)$ is continuous.

Example. Using the product rule, differentiate $(x^2+2x-1)(x^3-4x^2)$.

Solution. \begin{eqnarray*}\frac{d}{dx}[(x^2+2x-1)(x^3-4x^2)]&=&\frac{d(x^2+2x-1)}{dx}(x^3-4x^2)+(x^2+2x-1)\\&&\frac{d(x^3-4x^2)}{dx}\\&=&(2x+2)(x^3-4x^2)+(x^2+2x-1)(3x^2-8x)\\&=&(2x^4-6x^3-8x^2)+(3x^4-2x^3-19x^2+8x)\\&=&5x^4-8x^3-27x^2+8x.\end{eqnarray*} Multiplyng first, \begin{eqnarray*}(x^2+2x-1)(x^3-4x^2)&=&x^5-4x^4+2x^4-8x^3-x^3+4x^2\\&=&x^5-2x^4-9x^3+4x^2.\end{eqnarray*} The derivative of this is $5x^4-8x^3-27x^2+8x$ by the power rule and differentiation formulas we discussed here.

Reciprocal Rule (Baby Quotient Rule): Let $v=g(x)$ be a differentiable function with $g(x)\ne 0$. Then $$\left(\frac{1}{g}\right)’(x)=\frac{-g’(x)}{[g(x)]^2}$$ or $$\frac{d}{dx}\left(\frac{1}{v}\right)=-\frac{1}{v^2}\frac{dv}{dx}.$$

Proof. \begin{eqnarray*}\left(\frac{1}{g}\right)’(x)&=&\lim_{\Delta x\to 0}\frac{\frac{1}{g(x+\Delta x)}-\frac{1}{g(x)}}{\Delta x}\\&=&\lim_{\Delta x\to 0}\frac{\frac{g(x)-g(x+\Delta x)}{g(x+\Delta x)g(x)}}{\Delta x}\\&=&-\lim_{\Delta x\to 0}\frac{\frac{g(x+\Delta x)-g(x)}{\Delta x}}{g(x+\Delta x)g(x)}\\&=&-\frac{g’(x)}{[g(x)]^2}.\end{eqnarray*}

Example. Differentiate $\frac{1}{\sqrt{x}+2}$.

Solution. \begin{eqnarray*}\frac{d}{dx}\frac{1}{\sqrt{x}+2}&=&-\frac{d(\sqrt{x}+2)/dx}{(\sqrt{x}+2)^2}\\&=&-\frac{1}{2\sqrt{x}(\sqrt{x}+2)^2}.\end{eqnarray*}

Using the product rule and the reciprocal rule, we can prove

Quotient Rule: Let $u=f(x)$ and $v=g(x)$ be differentiable functions and assume that $g(x)\ne 0$. Then $$\left(\frac{f}{g}\right)’(x)=\frac{f’(x)g(x)-f(x)g’(x)}{[g(x)]^2}$$ or $$\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{\frac{du}{dx}v-u\frac{dv}{dx}}{v^2}.$$

Proof. \begin{eqnarray*}\left(\frac{f}{g}\right)’(x)&=&\left(f\frac{1}{g}\right)’(x)\\&=&f’(x)\frac{1}{g(x)}+f(x)\left(\frac{1}{g}\right)’(x)\ (\mbox{the product rule is applied})\\&=&\frac{f’(x)}{g(x)}-f(x)\frac{g’(x)}{[g(x)]^2}\ (\mbox{the reciprocal rule is applied})\\&=&\frac{f’(x)g(x)-f(x)g’(x)}{[g(x)]^2}.\end{eqnarray*}

Example. Find the derivative of $h(x)=\frac{2x+1}{x^2-2}$.

Solution. \begin{eqnarray*}h’(x)&=&\frac{(2x+1)’(x^2-2)-(2x+1)(x^2-2)’}{(x^2-2)^2}\\&=&\frac{2(x^2-2)-(2x+1)2x}{(x^2-2)^2}\\&=&\frac{2x^2-4-4x^2-2x}{(x^2-2)^2}\\&=&-\frac{2x^2+2x+4}{(x^2-2)^2}.\end{eqnarray*}

Example. Differentiate $\frac{x^2+2}{x^8}$.

Solution. Since the function is a rational function, you may hastily try to use the quotient rule to differentiate it. There is nothing wrong with that except there may be a simpler way to differentiate the function. In fact the function can be written as $$\frac{x^2+2}{x^8}=\frac{x^2}{x^8}+\frac{2}{x^8}=\frac{1}{x^6}+2x^{-8}=x^{-6}+2x^{-8}.$$ Thus the derivative is $$-6x^{-7}-16x^{-9}=-\frac{6}{x^7}-\frac{16}{x^9}.$$

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5 Responses to The Product and Quotient Rules

  1. subashp says:

    Nice examples of product rule. Thanks for sharing.

  2. Shouldn’t \(\frac{x^2}{x^8} = \frac{1}{x^6} \) and not \(\frac{1}{x^4}\)?

  3. Pingback: Basic Derivative Examples | web-trends.net

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