Derivative of Inverse of Hyperbolic

**BabyMilo** · June 18th 2011, 02:13 AM

Show that $\frac{d}{dx}(sinh^{-1}\sqrt{(x^2-1)}) = \frac{1}{\sqrt{(x^2-1)}}$

**Ackbeet** · June 18th 2011, 02:22 AM

What have you tried so far?

**BabyMilo** · June 18th 2011, 02:44 AM

Originally Posted by Ackbeet

What have you tried so far?

Sorry Solved it now.

Sorry

$sinh y = \sqrt{(x^2-1)}$
$cosh y \frac{dy}{dx} = \frac{x}{\sqrt{x^2-1}}$
$\frac{dy}{dx} = \frac{x}{\sqrt{x^2-1}\sqrt{1+(x^2-1)}}$ I originally didn't see $1 + (x^2-1) = x^2$
$\frac{dy}{dx} = \frac{x}{\sqrt{x^2-1}\sqrt{x^2}}$
$\frac{dy}{dx} = \frac{x}{x\sqrt{x^2-1}}$
$\frac{dy}{dx} = \frac{1}{\sqrt{x^2-1}}$

**Ackbeet** · June 18th 2011, 02:51 AM

Hmm, interesting. I actually don't agree with the result unless x is non-negative. If that was assumed in the problem statement, great. Otherwise, you can't simplify

$\sqrt{x^{2}}=x.$ You can say that $\sqrt{x^{2}}=|x|.$ So you could simplify as follows:

$\frac{x}{|x|\sqrt{x^{2}-1}}=\frac{\text{sgn}(x)}{\sqrt{x^{2}-1}},$

where sgn is the signum function.

Otherwise, your working is correct, I think.

**BabyMilo** · June 18th 2011, 03:34 AM

thank you...if so how do i approach the question?

**Ackbeet** · June 18th 2011, 03:43 AM

Just present the proof you had in Post # 3, with the modifications mentioned in Post # 4 and Post # 4's answer. If the problem originally stated that x is positive, you can simplify the signum to the expression in the OP. Otherwise, I think you have to leave it in there. To be on the safe side, include a sentence explaining what you're doing.

**Prove It** · June 18th 2011, 05:20 AM

Originally Posted by BabyMilo

Show that $\frac{d}{dx}(sinh^{-1}\sqrt{(x^2-1)}) = \frac{1}{\sqrt{(x^2-1)}}$

You could always convert the inverse hyperbolic sine function to its logarithmic equivalent...

$\displaystyle \begin{align*} y &= \sinh^{-1}{(x)} \\ \sinh{(y)} &= x \\ \frac{e^y - e^{-y}}{2} &= x \\ e^y - \frac{1}{e^y} &= 2x \\ e^{2y} - 1 &= 2x\,e^y \\ e^{2y} - 2x\,e^y - 1 &= 0 \\ e^{2y} - 2x\,e^y + \left(-x\right)^2 - \left(-x\right)^2 - 1 &= 0 \\ \left(e^y - x\right)^2 - x^2 - 1 &= 0 \\ \left(e^y - x\right)^2 &= x^2 + 1 \\ e^y - x &= \pm \sqrt{x^2 + 1} \\ e^y &= x \pm \sqrt{x^2 + 1} \\ y &= \ln{\left(x \pm \sqrt{x^2 + 1}\right)} \end{align*}$

Of course, since $\displaystyle \sqrt{x^2 + 1} > x$ for all $\displaystyle x$ , that means we can only accept $\displaystyle y = \ln{\left(x + \sqrt{x^2 + 1}\right)}$ .

Therefore

$\displaystyle \begin{align*}y &= \sinh^{-1}{\left(\sqrt{x^2 - 1}\right)} \\ y &= \ln{\left(\sqrt{x^2 - 1} + \sqrt{\left(\sqrt{x^2 - 1}\right)^2 + 1}\right)} \\ y &= \ln{\left(\sqrt{x^2 - 1} + \sqrt{x^2}\right)}\end{align*}$

You can now find the derivative using the chain rule, or exponentiating both sides and using implicit differentiation.

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