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