`y = ln((1 + e^x)/(1 - e^x))` Find the derivative.

Wednesday, March 25, 2015

$\displaystyle{y}={\ln{{\left(\frac{{{1}+{{e}}^{{x}}}}{{{1}-{{e}}^{{x}}}}\right)}}}$ Find the derivative.

Find the derivative of $\displaystyle{y}={\ln{{\left(\frac{{{1}+{{e}}^{{x}}}}{{{1}-{{e}}^{{x}}}}\right)}}}$ :

Use a property of the natural logarithm to rewrite as:

$\displaystyle{y}={\ln{{\left({1}+{{e}}^{{x}}\right)}}}-{\ln{{\left({1}-{{e}}^{{x}}\right)}}}$

If u is a differentiable function of x, then $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\ln{{\left({u}\right)}}}=\frac{{{d}{u}}}{{u}}$ so

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}=\frac{{{e}}^{{x}}}{{{1}+{{e}}^{{x}}}}-\frac{{-{{e}}^{{x}}}}{{{1}-{{e}}^{{x}}}}$

Subtracting the fractions we get:

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}=\frac{{{{e}}^{{x}}{\left({1}-{{e}}^{{x}}\right)}-{\left(-{{e}}^{{x}}\right)}{\left({1}+{{e}}^{{x}}\right)}}}{{{\left({1}+{{e}}^{{x}}\right)}{\left({1}-{{e}}^{{x}}\right)}}}$

Clearing the parantheses and adding like terms we get:

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}=\frac{{{2}{{e}}^{{x}}}}{{{1}-{{e}}^{{{2}{x}}}}}$

Ninenotes

Wednesday, March 25, 2015

$\displaystyle{y}={\ln{{\left(\frac{{{1}+{{e}}^{{x}}}}{{{1}-{{e}}^{{x}}}}\right)}}}$ Find the derivative.

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