how do you find the derivative for (e^x-e^-x)/(e^x+e^-x)?

Question

How do you find the derivative for this question?
(e^x-e^-x)/(e^x+e^-x)
I cant find the correct answer.. Im just not sure how to do it? I've been trying to use the quotient rule - f'g-fg'/g^2 but i cant seem to get it to work.

Answer 1

.......e^x-e^-x

y = -----------------

......e^x+e^-x



............(e^x+e^-x) d/dx e^(x) + e^(-x) - [e^x-e^-x] d/dx e^(x) - e^(-x)

dy/dx =---------------------------------------------------------------------------------------

.............................(e^x+e^-x)^2



.....[e^(x) + e^(-x)][e^(x)+e^(-x)] - [e^(x) - e^(-x)] [e^(x) -e^(-x) ]

=-------------------------------------------------------------------------------------

..........................(e^x+e^-x)^2



.......e^(2x) + e^(0) + e^(0) + e^(-2x) - [e^(2x) - e^(0) - e^(0) + e^(-2x)]

=---------------------------------------------------------------------------------------------------

........................(e^x+e^-x)^2



.......e^(2x) + 2 + e^(-2x) - [e^(2x) - 2 + e^(-2x)]

=---------------------------------------------------------------------

.....................(e^x+e^-x)^2



.......e^(2x) + 2 + e^(-2x) - e^(2x) + 2 - e^(-2x)

=-------------------------------------------------------------------

..................(e^x+e^-x)^2



...........4

= -----------------

...(e^x+1/e^(x))^2



.........4

=-----------------

....[e^(2x) + 1]^2

...-----------------

.........e^(2x)



..........e^(2x)

= 4 * -----------

.........[e^(2x) + 1]^2



.......4e^(2x)

=------------------------ answer//

......[e^(2x) + 1]^2

Answer 2

First eliminate all the negative indices by multiplying the fraction by (e^x) / (e^x):



y = [e^(2x) - 1] / [e^(2x) + 1]



Now rewrite the fraction as:



y = [e^(2x) + 1] / [e^(2x) + 1] - 2 / [e^(2x) + 1]

y = 1 - 2[e^(2x) - 1]^-1



Differentiate using the chain rule:



y' = 2[e^(2x) - 1]^-2 * e^(2x) * 2

y' = 4e^(2x) / [e^(2x) - 1]^2

Answer 3

u(x) = e^x - e^-x

u'(x) = e^x - [-e^-x]

u'(x) = e^x+e^-x



v(x) = e^x+e^-x

v'(x) = e^x-e^-x



d/dx [(e^x-e^-x)/(e^x+e^-x)] = d/dx [ u(x) / v(x)]



= ( v(x) u'(x) - u(x) v'(x) ) / ( v(x) )^2



= [(e^x+e^-x) (e^x+e^-x) - (e^x - e^-x)( e^x-e^-x) ] / ( e^x+e^-x)^2

= [(e^(2x)+1+1+e^(-2x)) - (e^(2x)-1-1+e^(-2x) )] / ( e^x+e^-x)^2

= ( e^(2x) + 2 + e^(-2x) -e^(2x)+2 -e^(-2x) ) / ( e^x+e^-x)^2

= 4 / ( e^x+e^-x)^2

Answer 4

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

y = (e^x-1/e^x)/(e^x+1/e^x)

y = (e^2x-1)/(e^2x+1)

dy/dx = [(e^2x+1)(e^2x.2)-(e^2x-1)(e^2x.2)]/(e^2x+1)^2

dy/dx = [2e^4x+2e^2x-2e^4x+2e^2x]/(e^2x+1)^2

dy/dx = 4e^2x/(e^2x+1)^2

dy/dx = 4e^2x/[e^x(e^x+1/e^x)]^2

dy/dx = 4e^2x/[e^2x(e^x+e^-x)^2]

dy/dx = 4/(e^x+e^-x)^2