Question

LGBADERIVATIVE LEVEL 2:

\[\huge y = \frac{e^x - e^{-x}}{e^x + e^{-x}}\]

Answer 1

okay..this time idk how to do this but it looks like sinh or cosh idk

Answer 2

\[y=\tanh(x)\]

Answer 3

how?

Answer 4

by definition

I didn't give you the answer..just telling what your original equation is equivalent to

Answer 5

yeah i know but how is that thingy be tanh

Answer 6

\[\tanh(x)=\frac{\sinh(x)}{\cosh(x)}\]

Answer 7

could it be that \[e^x - e^{-x} \implies \sinh (x)\]
\[e^x + e^{-x} \implies cosh (x)?\]

Answer 8

\[\sinh(x)=\frac{e^x-e^{-x}}{2}\]

Answer 9

ohh hmm anyway regarding this...

\[\large \frac{d}{dx} \tanh (x) = \ln |\cosh (x)|?\]

Answer 10

are you integrating or differentiating?

Answer 11

uhh oops lol

Answer 12

so it would be
\[\Large \frac{\cosh ^2 (x) - \sinh^2 (x) }{\cosh^2 (x)}\]
that right??

Answer 13

yes...or you can just write \(\text{sech}^2(x)\)

Answer 14

\[\huge 1 - \tanh^2 (x) = \sec h^2 (x)?\]

Answer 15

yes

Answer 16

okay thanks! :D