What is the derivative of $ln (\frac{t}{t^{2}})$ $ln( t /t^2)$ ?

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Answer 1 · 2015-10-18T17:59:18

$- \frac{1}{t}$ $-1/t$

Using the chain rule together with other standard rules of differentiation and laws of exponents and logs, we get:

$\frac{d}{d t} ln (\frac{t}{t^{2}}) = \frac{d}{d t} ln (\frac{1}{t})$ $d/dtln(t/t^2)=d/dtln(1/t)$

$= \frac{1}{\frac{1}{t}} \cdot \frac{d}{d t} (\frac{1}{t})$ $=1/(1/t)*d/dt(1/t)$

$= t \cdot (- t^{- 2})$ $=t*(-t^(-2))$

$= - \frac{1}{t}$ $=-1/t$

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