Question

find and simplify the derivatives of
f(t)=cosh^2(3t) - sinh^2(3t)

Answer 1

trick question

Answer 2

must be some more of that "trickonometry" that myininaya likes

Answer 3

This should help: http://www.math.com/tables/derivatives/tableof.htm

Answer 4

no it is a trick

Answer 5

recall:
\[\cosh(t)=\frac{e^t+e^{-t}}{2}\]
and
\[\sinh(t)=\frac{e^t-e^{-t}}{2}\]

Answer 6

oh c'mon...

Answer 7

...

Answer 8

\[\cosh(3t)=\frac{e^{3t}+e^{-3t}}{2}\]
and
\[\sinh(3t)=\frac{e^{3t}-e^{-3t}}{2}\]
--------------------------------------
\[([\cosh(3t)]^2)'=2\cosh(3t) \cdot \frac{3e^{3t}-3e^{-3t}}{2}=2\cdot \frac{e^{3t}+e^{-3t}}{2} \cdot \frac{3e^{3t}-3e^{-3t}}{2}\]

Answer 9

what on earth. ok i guess i have to spill the beans

Answer 10

i don't remember extra formulas

Answer 11

theres no point

Answer 12

supposed you were asked to find the derivative of
\[\sin^2(3x)+\cos^2(3x)\] what would you get?

Answer 13

yes i know the derivative of the thing is 0

Answer 14

why you would get zero right? because the most basicest trig identity is
\[\sin^2(x)+\cos^2(x)=1\]

Answer 15

but wouldn't it be cooler to do all that algebra

Answer 16

and the analogous one for hyperbolic functions is
\[\cosh^2(x)-\sinh^2(x)=1\] so guess what the derivative is...

Answer 17

so the answer is zero?

Answer 18

cooler if you think it would be cooler to find the derivative of
\[\sin^2(x)+\cos^2(x)=1\] by doing algebra