How do I differentiate the natural logarithm? For example, y=3ln(6t+1)

Wednesday, March 23, 2011

How do I differentiate the natural logarithm? For example, y=3ln(6t+1)

$\displaystyle{y}={3}{\ln{{\left({6}{t}+{1}\right)}}}$

To take the derivative of this, refer to formula:

$\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({\ln{{u}}}\right)}=\frac{{1}}{{u}}\cdot\frac{{{d}{u}}}{{\left.{d}{x}}}$

Applying that, the derivative of the function will be:

$\displaystyle\frac{{d}}{{{\left.{d}{t}\right.}}}{\left({y}\right)}=\frac{{d}}{{{\left.{d}{t}\right.}}}{\left[{3}{\ln{{\left({6}{t}+{1}\right)}}}\right]}$

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}={3}\frac{{d}}{{{\left.{d}{t}\right.}}}{\left[{\ln{{\left({6}{t}+{1}\right)}}}\right]}$

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}={3}\cdot\frac{{1}}{{{6}{t}+{1}}}\cdot\frac{{d}}{{{\left.{d}{t}\right.}}}{\left({6}{t}+{1}\right)}$

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}={3}\cdot\frac{{1}}{{{6}{t}+{1}}}\cdot{6}$

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}=\frac{{18}}{{{6}{t}+{1}}}$

Therefore, $\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}=\frac{{18}}{{{6}{t}+{1}}}$ .

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Wednesday, March 23, 2011