# Thread: need help dervative of hyperbolic

1. ## need help dervative of hyperbolic

Find d/dx [arcsinh (x)]

2. the derivative of archsinh(x) is $\displaystyle \dfrac{1} {\sqrt{x^{2}+1}}$

3. Hello, gracy!

Viet is correct . . . There is a formula for it.

Find $\displaystyle \frac{d}{dx}\left[\text{arcsinh}(x)\right]$
If we are expected to derive that formula, we need to know a few facts:

. . $\displaystyle \frac{d}{dx}\left(\sinh(y)\right) \:=\:\cosh(y)$ [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!