1. ## hyperbolic function derivatives!!

ok so how exactly would I simplify this function

sinh(ln(t))

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

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

-Dan

3. Before takin' the derivative, observe that

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

4. 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??

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

$\sinh x = \frac{{e^x - e^{ - x} }}
{2}.$
Now contemplate $x=\ln t$ and take its derivative.
Ha! I never saw that.

-Dan

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

$\sinh x = \frac{{e^x - e^{ - x} }}
{2}.$
Now contemplate $x=\ln t$ and take its derivative.
Nice, so the derivative is just 1?

7. Originally Posted by colby2152
Nice, so the derivative is just 1?
No.

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

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

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

-Dan

8. Originally Posted by topsquark
No.

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

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

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

-Dan
Whoops, gotta watch where I throw those negative signs!