Find d/dx [arcsinh (x)]
Hello, gracy!
Viet is correct . . . There is a formula for it.
If we are expected to derive that formula, we need to know a few facts:Find $\displaystyle \frac{d}{dx}\left[\text{arcsinh}(x)\right] $
. . $\displaystyle \frac{d}{dx}\left(\sinh(y)\right) \:=\:\cosh(y)$ [1]
. . $\displaystyle \cosh^2(y) - \sinh^2(y)\:=\:1\quad\Rightarrow\quad\cosh(y) \:=\:\sqrt{1 + \sinh^2(y)} $ [2]
We have: .$\displaystyle y \;=\;\text{arcsinh}(x)$
Then: .$\displaystyle \sinh(y) \;=\;x$ [3]
Using [1], differentiate implicitly: .$\displaystyle \cosh(y)\cdot\frac{dy}{dx} \;=\;1$
. . Then we have: .$\displaystyle \frac{dy}{dx}\:=\:\frac{1}{\cosh(y)}$
From [2], we have: .$\displaystyle \frac{dy}{dx}\:=\:\frac{1}{\sqrt{1 + \sinh^2(y)}} $
From [3], we have: .$\displaystyle \frac{dy}{dx}\:=\:\frac{1}{\sqrt{1 + x^2}}$ . . . ta-DAA!