deriving the formula

**Hongosh** · March 9th 2009, 03:25 PM

derive the formula d(sinh^-1 z)/dz=1/(1+z^2)^1/2 also i know there needs to be certain conditions that make this possible

**manyarrows** · March 9th 2009, 05:59 PM

I was just doing these for class last unit. You have to use the formula for deriving Inverse functions 1/f ' (g(x)) where f(x) is the inverse of g(x). In this case f (x)=sinch x and g(x)=arcsinch x (I don't know if it is legal to call it arcsinch, but I don't know how to write the math formulas on the computer) Anyway after this use the identity
cosh x= (1-(sinch x)^2)^1/2). try it from there and keep in mind it is similar to deriving inverse trig functions.

**mr fantastic** · March 9th 2009, 08:24 PM

Originally Posted by Hongosh

derive the formula d(sinh^-1 z)/dz=1/(1+z^2)^1/2 also i know there needs to be certain conditions that make this possible

$y = \sinh^{-1} z \Rightarrow z = \sinh y$ .

Therefore $\frac{dz}{dy} = \cosh y = \sqrt{1 + \sinh^2 y} = \sqrt{1 + z^2}$ .

You should think about why it's $\sqrt{1 + \sinh^2 y}$ and not $-\sqrt{1 + \sinh^2 y}$ .

Therefore $\frac{dy}{dz} = \frac{1}{\sqrt{1 + z^2}}$ .

**Soroban** · March 10th 2009, 05:56 AM

Hello, Hongosh!

A variation of Mr. F's solution . . .

Derive the formula: . $\frac{d}{dx}(\sinh^{-1}\!z) \;=\;\frac{1}{\sqrt{1+z^2}}$

We have: . $y \:=\:\sin^{-1}\!z \quad\Rightarrow \quad \sinh y \:=\:z$ .[1]

Differentiate implicitly: . $\cosh y\cdot\frac{dy}{dz} \:=\:1 \quad\Rightarrow\quad \frac{dy}{dz} \:=\:\frac{1}{\cosh y}$ .[2]

$\text{Since }\:\cosh^2\!y - \sin^2\!y \:=\:1\quad\Rightarrow\quad \cosh y\:=\:\sqrt{1 + \sinh^2\!y}$

Substitute [1]: . $\cosh y \:=\:\sqrt{1+z^2}$

Substitute into [2]: . $\frac{dy}{dx} \;=\;\frac{1}{\sqrt{1+z^2}}$

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