Hyperbolic Trig: Solutions

y = e^sinh²x
$dy dx &sp;=&sp;
2 (sinh x) (cosh x)
(e$ ^sinh²x)
$y &sp;=&sp; tanh x x$
$dy dx &sp;=&sp; x&sp; sech^2^x
- tanh x x^2^$
y = cosh(e^lnx)
$dy dx &sp;=&sp; sinh (e^ln x^) e^ln x^x &sp;=&sp; sinh x$
This should seem obvious since e^lnx = x

jjnichol@mit.edu