# Hyperbolic Diff helppp

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• Jan 18th 2007, 05:55 AM
^_^Engineer_Adam^_^
Hyperbolic Diff helppp
Hello :)
Find the derivative of the function
$g(x)=ln(tanh(x))$

why is the answer
$2csch(2x)$?
• Jan 18th 2007, 06:27 AM
ThePerfectHacker
Quote:

Originally Posted by ^_^Engineer_Adam^_^
Hello :)
Find the derivative of the function
$g(x)=ln(tanh(x))$

why is the answer
$2csch(2x)$?

You use the chain rule.
$y=\ln u$ where $u=\tanh x$
$dy/du=1/u=1/\tanh x=\mbox{coth} x$
$du/dx=\mbox{sech}^2 x$
Thus,
$dy/dx=\mbox{sech}^2 x\cdot \mbox{coth}x$
Express as fractions and simiplify to get,
$\frac{1}{\sinh x\cosh x}$
Multiply numerator and denoimator by 2,
$\frac{2}{2\sinh x\cosh x}$
Double angle for hyperbolic functions,
$\frac{2}{\sinh 2x}=2\mbox{csch} 2x$
• Jan 18th 2007, 06:45 AM
Soroban
Hello, ^_^Engineer_Adam^_^!

Quote:

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

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

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

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

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

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

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

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

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