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

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- Sep 10th 2007, 04:26 AMelbartoInverse 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 - Sep 10th 2007, 04:50 AMtopsquark
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 - Sep 10th 2007, 05:08 AMelbarto
Hi dan,

Im just trying to find the derivative of y. is it as simple as

$\displaystyle y^{\prime} = ar~cosh(2x) \cdot 2 = 2ar~cosh(2x)$?

Thanks for your help

- Sep 10th 2007, 05:28 AMSoroban
Hello, Elbarto!

There is a formula for this problem.

If you don't know it, we can derive it.

Quote:

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

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: .$\displaystyle \frac{dy}{dx}\;=\;\frac{4}{\sqrt{1 + (4x)^2}} \;=\;\frac{4}{\sqrt{1+16x^2}}$

- Sep 10th 2007, 09:24 AMtopsquark
Oh! That "ar" on the outside of the sinh(x), that was supposed to mean "arcsinh(x)"? Sorry about that.

-Dan