Question

LGBADERIVATIVES:

\[\huge G(x) = \frac{1-\cosh x}{1+\cosh x}\]

Answer 1

Use the division rule, it's pretty easy

Answer 2

quotient rule immediately @ash2326 ? cant it be easier???

Answer 3

it looks intimidating

Answer 4

\[\huge \frac{\sinh x(1+\cosh x) - \sinh x(1-\cosh x)}{(1+\cosh x)^2}?\]

Answer 5

It's easy
since
\[\frac{d}{dx} (\cos hx)=\sinh x \]

\[\frac{(1+\cosh x)\times (-\sinh x)-(1-\cosh x)\times (\sinh x)}{(1+\cosh x)^2} \]
just simplification now

Answer 6

\[\huge \frac{\sinh x(2\cosh x)}{(1+\cosh x)^2}?\]

Answer 7

just
\[\frac{-2\sinh x}{(1+\cosh x)^2}\]

Answer 8

why negative?

Answer 9

oh wait yes

Answer 10

ha ha

Answer 11

i forgot it was 1-coshx