1. ## Hyperbolic Diff helppp

Hello
Find the derivative of the function
$\displaystyle g(x)=ln(tanh(x))$

$\displaystyle 2csch(2x)$?

Hello
Find the derivative of the function
$\displaystyle g(x)=ln(tanh(x))$

$\displaystyle 2csch(2x)$?
You use the chain rule.
$\displaystyle y=\ln u$ where $\displaystyle u=\tanh x$
$\displaystyle dy/du=1/u=1/\tanh x=\mbox{coth} x$
$\displaystyle du/dx=\mbox{sech}^2 x$
Thus,
$\displaystyle dy/dx=\mbox{sech}^2 x\cdot \mbox{coth}x$
Express as fractions and simiplify to get,
$\displaystyle \frac{1}{\sinh x\cosh x}$
Multiply numerator and denoimator by 2,
$\displaystyle \frac{2}{2\sinh x\cosh x}$
Double angle for hyperbolic functions,
$\displaystyle \frac{2}{\sinh 2x}=2\mbox{csch} 2x$

Find the derivative of the function: .$\displaystyle g(x) \:=\:\ln[\text{tanh}(x)]$

Why is the answer: .$\displaystyle g'(x) \:=\:2\,\text{csch}(2x)$ ?

You're expected to know these facts for hyperbolic functions:

. . $\displaystyle \frac{d}{dx}[\text{tanh}(x)] \:=\:\text{sech}^2(x)$

. . $\displaystyle \text{tanh}(x) \:=\:\frac{\text{sinh}(x)}{\text{cosh}(x)}\qquad\t ext{sech}(x)\:=\:\frac{1}{\text{cosh}(x)}\qquad\te xt{csc}(x) \:=\:\frac{1}{\text{sinh}(x)}$

. . $\displaystyle 2\,\text{sinh}(x)\text{cosh}(x) \:=\:\text{sinh}(2x)$

We have: .$\displaystyle g(x)\:=\:\ln[\text{tanh}(x)]$

Then: .$\displaystyle g'(x) \:=\:\frac{1}{\text{tanh}(x)}\cdot\text{sech}^2(x) \:=\:\frac{1}{\frac{\text{sinh}(x)}{\text{cosh}(x) }}\cdot\frac{1}{\text{cosh}^2(x)} \:=\:\frac{1}{\text{sinh}(x)\text{cosh}(x)}$

Multiply top and bottom by 2: .$\displaystyle \frac{2}{2\,\text{sinh}(x)\text{cosh}(x)} \;=\;\frac{2}{\text{sinh}(2x)} \;=\;2\,\text{csch}(2x)$