Comparing Hyperbolic and Cartesian Trig Properties

[chwala]

Homework Statement

See attached.

Relevant Equations

hyperbolic trig. properties

I came across this question; i noted that the hyperbolic trigonometry properties are somewhat similar to what i may call cartesian trigonometry properties...

My approach on this;

$\tanh x = \sinh y$

...just follows from

$y=\sin^{-1}(\tan x)$

$\tan x = \sin y$

Therefore continuing with our problem;

$\sech^{2}x= \cosh y \dfrac{dy}{dx}$

$⇒\dfrac{dy}{dx}= \dfrac{\sech^{2}x}{\cosh y}$

We know that;

$\cosh^2 y - \sinh^2y =1$

Therefore,

$\dfrac{dy}{dx}= \dfrac{\sech^{2}x}{\sqrt{1+\sinh^2y}}$

which gives us;

$\dfrac{dy}{dx}= \dfrac{\sech^{2}x}{\sqrt{1+\tanh^2x}}$

would appreciate insight or any other approach...

 

[vanhees71]

Why don't you simply use the chain rule?