Differentiate #\frac{e^{8u}-e^{-8u}}{e^{8u}+e^{-8u}}# ?

Question

1905 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-14T07:08:17

#f'(u)=(32*e^(16u))/(1+e^(16u))^2#

Multiplying numerator and denominator by

#e^(8u)#
we get
#f(u)=(e^(16u)-1)/(e^(16u)+1)#
now we get by the Quotient and chain rule

#f'(u)=((e^(16u)*16*(e^(16u)+1)-(e^(16u)-1)*e^(16u)*16))/(e^(16u)+1)^2#
simplifying we get the result above.

Answer 2 · 2018-06-14T09:18:19

Please see the explanation below.

We know that

#(e^(8u)-e^(-8u))/(e^(8u)+e^(8u))=tanh(8u)#

Therefore, the derivative is

#(tanh(8u))'=8sech^2(8u)#

#=8/cosh^2(8u)#

#=32/(e^(8u)+e^(-8u))^2#

#=(32e^(16u))/(e^(16u)+1)^2#