# Thread: Inverse Hyperbolic Trig Functions?

1. ## Inverse Hyperbolic Trig Functions?

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

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

-Dan

3. Hi dan,

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

4. Hello, Elbarto!

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

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

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

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

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

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

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

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

-Dan