differentiation

**stealthmaths** · March 15th 2010, 02:35 PM

Hi:
I have a complicated question here. At least, I think so.
I have gotten to a stage, but not sure whether I needed to do the last stage. Did I need to get y on its own or was the differentiation fine.
And now, do I just differentiate again and then put the answers of the two into the equation provided to see if it is equal to zero?

**skeeter** · March 15th 2010, 02:48 PM

Originally Posted by stealthmaths

Hi:
I have a complicated question here. At least, I think so.
I have gotten to a stage, but not sure whether I needed to do the last stage. Did I need to get y on its own or was the differentiation fine.
And now, do I just differentiate again and then put the answers of the two into the equation provided to see if it is equal to zero?

$\frac{d}{dx}(e^y = e^x + e^{-x})$

$e^y \cdot \frac{dy}{dx} = e^x - e^{-x}<br />$

$\frac{dy}{dx} = \frac{e^x - e^{-x}}{e^y}<br />$

$\frac{d}{dx}\left[\frac{dy}{dx} = \frac{e^x - e^{-x}}{e^y}\right]$

$\frac{d^2y}{dx^2} = \frac{e^y(e^x + e^{-x}) - (e^x - e^{-x})e^y \cdot \frac{dy}{dx}}{e^{2y}}<br />$

$\frac{d^2y}{dx^2} = \frac{e^{2y} - (e^x - e^{-x})^2}{e^{2y}}<br />$

$\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 - 1 =$

$\frac{e^{2y} - (e^x - e^{-x})^2}{e^{2y}} + \frac{(e^x - e^{-x})^2}{e^{2y}} - 1 =$

$\frac{e^{2y}}{e^{2y}} - 1 = 0$

**stealthmaths** · March 15th 2010, 02:54 PM

Yeah that is a tricky problem. Well, doesn't seem so for you but...
Thanks skeeter. I'm gonna go over that one after I have done some more easier practice, I think.

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