Inverse tan and inverse of tanh

• Oct 31st 2010, 05:18 AM
MattWT
Inverse tan and inverse of tanh
Hello,

I know what the inverse of tan is when differentiated, but would the inverse of tanh differentiated follow the same format?

$\displaystyle -1/(1-x^2)$

Thank you!
• Oct 31st 2010, 05:23 AM
Prove It
Let's see...

$\displaystyle \displaystyle y = \tanh{x}$

$\displaystyle \displaystyle = \frac{\sinh{x}}{\cosh{x}}$

$\displaystyle \displaystyle \frac{dy}{dx} = \frac{\cosh{x}\,\frac{d}{dx}(\sinh{x}) - \sinh{x}\,\frac{d}{dx}(\cosh{x})}{\cosh^2{x}}$

$\displaystyle \displaystyle = \frac{\cosh{x}\cosh{x} - \sinh{x}\sinh{x}}{\cosh^2{x}}$

$\displaystyle \displaystyle = \frac{\cosh^2{x} - \sinh^2{x}}{\cosh^2{x}}$

$\displaystyle \displaystyle = \frac{1}{\cosh^2{x}}$

$\displaystyle \displaystyle = \textrm{sech}^2{x}$.

This is almost identical to $\displaystyle \displaystyle \frac{d}{dx}(\tan{x}) = \sec^2{x}$
• Oct 31st 2010, 06:20 AM
HallsofIvy
And therefore, with $\displaystyle x= tanh(y)$,
$\displaystyle d(tanh^{-1}(x))}{dx}= \frac{1}{sech^2(y)}= \frac{1}{1- tanh^2(y)}= \frac{1}{1- x^2}$