Product rule
Logarithmic differentiation

Let h (x) = f (x) g (x) {\displaystyle h(x)=f(x)g(x)}. Taking the absolute value of each function and the natural log of both sides of the equation,

Applying properties of the absolute value and logarithms,

Taking the logarithmic derivative of both sides and then solving for h ′ (x) {\displaystyle h'(x)}:

Solving for h ′ (x) {\displaystyle h'(x)} and substituting back f (x) g (x) {\displaystyle f(x)g(x)} for h (x) {\displaystyle h(x)} gives:

Note: 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 for positive arguments. This works because d d x (ln | u |) = u ′ u {\displaystyle {\tfrac {d}{dx}}(\ln |u|)={\tfrac {u'}{u}}}, which justifies taking the absolute value of the functions for logarithmic differentiation.
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