# hyperbolic function derivatives!!

• Mar 11th 2008, 10:33 AM
mathlete
hyperbolic function derivatives!!
ok so how exactly would I simplify this function

sinh(ln(t))
• Mar 11th 2008, 10:39 AM
topsquark
Quote:

Originally Posted by mathlete
ok so how exactly would I simplify this function

sinh(ln(t))

Use the chain rule:
$\displaystyle \frac{d}{dt}sinh(ln(t)) = cosh(ln(t)) \cdot \frac{1}{t}$

-Dan
• Mar 11th 2008, 10:43 AM
Krizalid
Before takin' the derivative, observe that

$\displaystyle \sinh x = \frac{{e^x - e^{ - x} }} {2}.$ Now contemplate $\displaystyle x=\ln t$ and take its derivative.
• Mar 11th 2008, 10:48 AM
mathlete
yeah i know how to take the derivative, but it asks me to simplify the function first...so to do that would i just plug in ln(t) into the (e^x-e^(-x))/2??
• Mar 11th 2008, 11:04 AM
topsquark
Quote:

Originally Posted by Krizalid
Before takin' the derivative, observe that

$\displaystyle \sinh x = \frac{{e^x - e^{ - x} }} {2}.$ Now contemplate $\displaystyle x=\ln t$ and take its derivative.

Ha! I never saw that. (Doh)

-Dan
• Mar 11th 2008, 11:06 AM
colby2152
Quote:

Originally Posted by Krizalid
Before takin' the derivative, observe that

$\displaystyle \sinh x = \frac{{e^x - e^{ - x} }} {2}.$ Now contemplate $\displaystyle x=\ln t$ and take its derivative.

Nice, so the derivative is just 1?
• Mar 11th 2008, 11:11 AM
topsquark
Quote:

Originally Posted by colby2152
Nice, so the derivative is just 1?

No.

$\displaystyle sinh(ln(x)) = \frac{e^{ln(x)} - e^{-ln(t)}}{2}$

$\displaystyle = \frac{x - \frac{1}{x}}{2}$

So the derivative will be
$\displaystyle \frac{1}{2} + \frac{1}{2x^2}$

-Dan
• Mar 11th 2008, 11:27 AM
colby2152
Quote:

Originally Posted by topsquark
No.

$\displaystyle sinh(ln(x)) = \frac{e^{ln(x)} - e^{-ln(t)}}{2}$

$\displaystyle = \frac{x - \frac{1}{x}}{2}$

So the derivative will be
$\displaystyle \frac{1}{2} + \frac{1}{2x^2}$

-Dan

Whoops, gotta watch where I throw those negative signs!