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Quotient rule

This is similar to the product rule except that it applies to quotients. We can deduce it by differentiating \( y = uv^{-1} \) using the product and chain rules. First, by the chain rule \(
\frac{\mathrm{d}}{\mathrm{d}x} v^{-1} = -v^{-2} \frac{\mathrm{d}v}{\mathrm{d}x} \). By the product rule,

\begin{eqnarray*}
\frac{\mathrm{d}y}{\mathrm{d}x} & = & v^{-1} \frac{\mathrm{d}u...
...thrm{d}u}{\mathrm{d}x} - u \frac{\mathrm{d}v}{\mathrm{d}x}}{v^2}
\end{eqnarray*}



This is the quotient rule used to differentiate a quotient of functions such as \( y = \frac{x}{\ln{x}} \). Here, we put \( u = x, v
= \ln{x} \), so that \( \frac{\mathrm{d}u}{\mathrm{d}x} = 1, \frac{\mathrm{d}v}{\mathrm{d}x} = \frac{1}{x}
\). Hence, \( \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\ln{x} - 1}{(\ln{x})^2} =
\frac{1}{\ln{x}} - \frac{1}{(\ln{x})^2} \).

Alexander Frolkin 2001-03-13