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suppose g(x)=|ln(x)| Find g'(1/2)
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Ok, with, |x| = sqrt((x)^2) = (x^2)^(1/2) We get, |ln(x)| = (ln(x))^2)^(1/2) We use chain rule to find g' = \[\frac{ 1 }{ 2 }(\ln(x)^{2})^{=1/2}*\left( 2\ln(x) * \frac{ 1 }{ x } \right) = \frac{ 1 }{ x }\] Which is also the derivative of just ln(x). Now using g'(1/2), we get 1/(1/2) = 2
ohh alright Thanks so much!
I believe the answer is -2
\[\frac{ 1 }{ 2 }(\ln(x)^{2})^{-1/2}\cdot\left( 2\ln(x) \cdot \frac{ 1 }{ x } \right) \] \[\left( \frac{\ln(x)}{(\ln(x)^{2})^{1/2}} \cdot \frac{ 1 }{ x } \right)\] \[\left( \frac{\ln(x)}{|\ln(x)|} \cdot \frac{ 1 }{ x } \right) = \text{sign}(\ln(x))\cdot\frac{ 1 }{ x }\]
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