1. ## Another Hyperbolic function

Given that, $sinh \ y = x$, show that $y = ln [x+(1+x^2)^\frac{1}{2}]$
By differentiating this result, show that

$(1+x^2)(\frac {dy}{dx})^2 = 1$

I’ve done first part, but how can I differentiating this result?

2. Hello,

Well, i think you got to differentiate $\sinh(\ln(x+(1+x^2)^{1/2}))$, which is sinh(y)=x. This derivate should be 1, because the derivate of x is 1.

For the derivate of $\sinh(\ln(x+(1+x^2)^{1/2}))$, use the fog derivate.
When you differentiate sinh, you got cosh. And cosh is $\sqrt{1+\sinh^2}$

3. Originally Posted by Moo
Hello,

Well, i think you got to differentiate $\sinh(\ln(x+(1+x^2)^{1/2}))$, which is sinh(y)=x. This derivate should be 1, because the derivate of x is 1.

For the derivate of $\sinh(\ln(x+(1+x^2)^{1/2}))$, use the fog derivate.
When you differentiate sinh, you got cosh. And cosh is $\sqrt{1+\sinh^2}$
I think sinh y = x is for first portion. Can you tell me the way how can I direct differentiate this? I tried by using ln x, dy/dx (xy).... etc. But can't solved yet.

4. Well,

We got sinh(y)=x.

So if we differentiate, we have :

$\frac{dy}{dx} \cosh(y) = \frac{d}{dx}x = 1$ (1)

(1)² will give : $(\frac{dy}{dx})^2 \cosh^2(y) = 1$

$\cosh^2(y)=1+\sinh^2(y)= 1+x^2$

And you obtain what you want :-)

5. Originally Posted by Moo
Well,

We got sinh(y)=x.

So if we differentiate, we have :

$\frac{dy}{dx} \cosh(y) = \frac{d}{dx}x = 1$ (1)

(1)² will give : $(\frac{dy}{dx})^2 \cosh^2(y) = 1$

$\cosh^2(y)=1+\sinh^2(y)= 1+x^2$

And you obtain what you want :-)
Look in my question, there are two parts,

first, Given that, sinh y = x, show that $y = ln [x+(1+x^2)^\frac{1}{2}]$

I solved that. No problem.

Second, By differentiating this result, show that

$(1+x^2)(\frac {dy}{dx})^2 = 1$

So why I start from sinh y = x?

6. Good luck then , i've tried to differentiate the expression of y, and it's far too complicated when the solution is here... I can't even know if i'll be able to get the result they're waiting from you

7. Originally Posted by geton
Given that, $sinh \ y = x$, show that $y = ln [x+(1+x^2)^\frac{1}{2}]$
By differentiating this result, show that

$(1+x^2)(\frac {dy}{dx})^2 = 1$

I’ve done first part, but how can I differentiating this result?
$y = \ln [x+(1+x^2)^{1/2}]$.

Let $u = x+(1+x^2)^{1/2} = x + \sqrt{1 + x^2}$ so that $y = \ln u$.

Chain rule: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{u} \cdot \frac{du}{dx}$.

To get $\frac{du}{dx}$ you need to differentiate $(1+x^2)^{1/2}$. So .... pretend for a moment that you have $y = (1+x^2)^{1/2}$:

Let $w = 1 + x^2$ so that $y = w^{1/2}$.

Chain rule: $\frac{dy}{dx} = \frac{dy}{dw} \cdot \frac{dw}{dx} = \frac{1}{2} \, w^{-1/2} \cdot 2x = \frac{x}{\sqrt{w}} = \frac{x}{\sqrt{1 + x^2}}$.

So $\frac{du}{dx} = 1 + \frac{x}{\sqrt{1 + x^2}} = \frac{\sqrt{1 + x^2} +x}{\sqrt{1 + x^2}}$.

Therefore:

$\frac{dy}{dx} = \frac{1}{u} \cdot \frac{du}{dx} = \frac{1}{x+ \sqrt{1+ x^2}} \cdot \frac{\sqrt{1 + x^2} +x}{\sqrt{1 + x^2}}$

$= \frac{\sqrt{1 + x^2} +x}{(x+ \sqrt{1+ x^2})\sqrt{1 + x^2}} = \frac{1}{\sqrt{1 + x^2}}$.

It should be all blue sky from here.