Inverse Hyperbolic Trig Functions?

**elbarto** · September 10th 2007, 05:26 AM

Hi,
Can somebody please help me with finding the derivative of this function? I can find the derivative if it is a normal function but the inverse function has me stumped at the moment.

Thanks

Elbarto

**topsquark** · September 10th 2007, 05:50 AM

Originally Posted by elbarto

Hi,
Can somebody please help me with finding the derivative of this function? I can find the derivative if it is a normal function but the inverse function has me stumped at the moment.

Thanks

Elbarto

Are you looking for the derivative of $ar~sinh(2x)$ or the derivative of its inverse?
$y^{\prime} = ar~cosh(2x) \cdot 2 = 2ar~cosh(2x)$
by the chain rule
and note that $\frac{d}{dx}sinh^{-1}(x) = \frac{1}{\sqrt{x^2 + 1}}$
so
$y = \frac{1}{2}sinh^{-1} \left ( \frac{x}{ar} \right )$
is the inverse function and has a derivative
$y^{\prime} = \frac{1}{2} \cdot \frac{1}{\sqrt{ \left ( \frac{x}{ar} \right )^2 + 1}} \cdot \frac{1}{ar} = \frac{1}{2\sqrt{x^2 + a^2r^2}}$
after some simplification.

-Dan

**elbarto** · September 10th 2007, 06:08 AM

Hi dan,

Im just trying to find the derivative of y. is it as simple as
$y^{\prime} = ar~cosh(2x) \cdot 2 = 2ar~cosh(2x)$ ?

Thanks for your help

**Soroban** · September 10th 2007, 06:28 AM

Hello, Elbarto!

There is a formula for this problem.
If you don't know it, we can derive it.

$y \;=\;\sinh^{-1}(4x)$

Take the $\sinh$ of both sides: . $\sinh y \;=\;4x$ .[1]

Differentiate implicitly: . $\cosh y\cdot\frac{dy}{dx}\;=\;4\quad\Rightarrow\quad\fra c{dy}{dx}\;=\;\frac{4}{\cosh y}<br />$ .[2]

From the identity: . $\cosh^2u - \sinh^2u \:=\:1$ , we have: . $\cosh u \:=\:\sqrt{1 + \sinh^2u}$

Then [2] becomes: . $\frac{dy}{dx} \;=\;\frac{4}{\sqrt{1+\sinh^2y}}$

Substitute [1] and we have: . $\frac{dy}{dx}\;=\;\frac{4}{\sqrt{1 + (4x)^2}} \;=\;\frac{4}{\sqrt{1+16x^2}}$

**topsquark** · September 10th 2007, 10:24 AM

Oh! That "ar" on the outside of the sinh(x), that was supposed to mean "arcsinh(x)"? Sorry about that.

-Dan

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