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L14 - Lecture 14 Derivative of Logarithm Functions...

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Lecture 14 — Derivative of Logarithm Functions; Logarithmic Differentiation Let a 6 = 1 be a positive number. d dx log a ( x ) = In particular d dx ln( x ) = From chain rule, we see d dx ln( g ( x )) = or in other words d dx ln( u ) = CAREFUL! What are the horizontal tangent lines of the graph of ln( x 2 + 5 x ) ?
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d dt ln(3 t - e - 3 t ) = d dx (ln( x )) 3 = d du ln(ln( u 2 + 5 u )) =
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Find d dx ln | x | An improvement: What are the horizontal tangent lines of the graph of ln | x 2 + 5 x | ?
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d dt log 3 | t 3 - 2 t | = d dx ln s e 2 x x 2 2 x - 1 =
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Logarithmic Differentiation A useful property of the logarithm function is that it turns products and quotients into sums, which are easier to differentiate. If an equation is especially complex, we can turn it into an implicit equation by taking the logarithm of each side and using these properties to our advantage. This process is called logarithmic differentiation. Find the slope of the tangent line to the graph of P ( x ) = x 1 / 3 e x (2 x - 1) 4 when x = 1 .
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Does logarithmic differentiation make sense when the function is negative? zero?
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