Hyperbolic Functions/Exponentials

I've been going through questions on OCR FP2 papers, and I've come across a problem with a few hyperbolic functions questions...

Show that the curve with equation: y = cosh 2x - 6 sinh x, has one stationary point and give its x coordinate in logarithmic form. Determine the nature of this stationary point.

Here's my working:

y = cosh 2x - 6 sinh x

dy/dx = 2 sinh 2x - 6 cosh x (d^2y/dx^2 = 4 cosh 2x - 6 sinh x)

dy/dx = e^2x - e^-2x - 3e^x - 3e^-x

Normally, I'd multiply through by e^x now to get a quadratic equation in e^x, and then solve for x, but I'm not too sure what to do with e^-2...

I tried multiplying by e^2x:

dy/dx = e^4x - 3e^3x - 3e^x - 1 = 0

If I try to take natural logs now, I end up with something like 8x = 0 (since ln 1 = 0), and its obvious that x isn't 0 in this case...

Any help would be great, I've come stuck on a few problems like this where you've got e^-2x included. :smile:

Reply 1

CalculusMan

Do you really need to use exponentials though?

dy/dx = 0
dy/dx = 2 sinh 2x - 6 cosh x = 0

2 sinh 2x - 6 cosh x = 0
sinh 2x - 3 cosh x = 0
2 sinh x cosh x - 3 cosh x = 0
cosh x(2 sinh x - 3) = 0

And go from there?

Reply 2

jazznaz

A teacher at my college led me to believe that it would be best to use the exponential forms (especially since the earlier part of the question asked you to derive the relation: sinh 2x = 2 sinh x cosh x using exponentials).

It does look easier just with the hyperbolic functions themselves though, I'll give it a go. Thanks :smile: