Editing Quotient rule (section)

=== Proof by logarithmic differentiation ===
Let <math>h(x)=\frac{f(x)}{g(x)}.</math> Taking the [[absolute value]] and [[natural logarithm]] of both sides of the equation gives
<math display="block">\ln|h(x)|=\ln\left|\frac{f(x)}{g(x)}\right|</math>

Applying properties of the absolute value and logarithms,
<math display="block">\ln|h(x)|=\ln|f(x)|-\ln|g(x)|</math>

Taking the [[logarithmic derivative]] of both sides,
<math display="block">\frac{h'(x)}{h(x)}=\frac{f'(x)}{f(x)}-\frac{g'(x)}{g(x)}</math> 

Solving for <math>h'(x)</math> and substituting back <math>\tfrac{f(x)}{g(x)}</math> for <math>h(x)</math> gives:
<math display="block">\begin{align}
h'(x)&=h(x)\left[\frac{f'(x)}{f(x)}-\frac{g'(x)}{g(x)}\right]\\
&=\frac{f(x)}{g(x)}\left[\frac{f'(x)}{f(x)}-\frac{g'(x)}{g(x)}\right]\\
&=\frac{f'(x)}{g(x)}-\frac{f(x)g'(x)}{g(x)^2}\\
&=\frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2}.
\end{align}</math>

Taking the absolute value of the functions is necessary for the [[logarithmic differentiation]] of functions that may have negative values, as logarithms are only [[Real-valued function|real-valued]] for positive arguments. This works because <math>\tfrac{d}{dx}(\ln|u|)=\tfrac{u'}{u}</math>, which justifies taking the absolute value of the functions for logarithmic differentiation.