Question

find y'

y=e^x+e^-x/e^x-e^-x

Answer 1

\[\Large y=\frac{e^x+e^{-x}}{e^x-e^{-x}}\]
@Andresfon12 do you know division rule of differentiation?

Answer 2

by using the quotant rule

Answer 3

yeah:)

Answer 4

so is y=(e^x+e^-x)'(e^x-e^-x)-(e^x+e^-x)(e^x-e^-x)'/(e^x-e^-x)^2

Answer 5

@ash2326

Answer 6

yeah, you're doing right

Answer 7

this is the hard part to find the top values

Answer 8

solve y = e^x+e^(-x)/e^x-e^(-x) for y
that equals
y = e^(-2 x)-e^(-x)+e^x

Answer 9

\[\large y=\frac{(e^x+e^{-x})'(e^x-e^{-x})-(e^x+e^{-x})(e^x-e^{-x})'}{(e^x-e^{-x})^2}\]

Answer 10

is it correct

Answer 11

u can see here clearly
http://www.wolframalpha.com/input/?i=find+y+if+y%3De%5Ex%2Be%5E-x%2Fe%5Ex-e%5E-x

Answer 12

now let's differentiate
\[\large y=\frac{(e^x-e^{-x})(e^x-e^{-x})-(e^x+e^{-x})(e^x+e^{-x})}
{(e^x-e^{-x})^2}\]
just simplification now
\[\large y=\frac{(e^x-e^{-x})^2-(e^x+e^{-x})^2}
{(e^x-e^{-x})^2}\]
Can you take it from here @Andresfone12?

Answer 13

chain rule afther that

Answer 14

nope, just simplification now

Answer 15

how im goin to do that @ash2326?

Answer 16

\[\large\frac{dy}{dx}=\frac{(e^x-e^{-x})^2-(e^x+e^{-x})^2}
{(e^x-e^{-x})^2}\]
it's of the form
\[a^2-b^2=(a+b)(a-b)\]
so
\[\large \frac{dy}{dx}=\frac{((e^x-e^{-x})+(e^x+e^{-x}))((e^{x}-e^{-x})-(e^x+e^{-x}))}
{(e^x-e^{-x})^2}\]
we get
\[\large \frac{dy}{dx}=\frac{(2e^x)(-2e^{-x})}{(e^x-e^{-x})^2}\]
we get finally
\[\large \frac{dy}{dx}=\frac{-4}{(e^x-e^{-x})^2}\]

Answer 17

do you get this?

Answer 18

still no over yet

Answer 19

it's over now

Answer 20

1/sinh^2x= csch^2x

Answer 21

@Andresfon12 you could use hyperbolic cos/sin and reduce it:)
Please try it and if you don't get I'd help you

Answer 22

so is the final ans is csch{^2}x when we use the hyp @ash2326?

Answer 23

with a minus sign

Answer 24

k