How do you differentiate #y=ln(e^x+sqrt(1+e^(2x)))#?

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Answer 1 · 2016-08-09T18:14:04

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

#u(x) = e^x + (1+e^(2x))^(1/2) and y = ln(u)#

#(dy)/(du) = 1/u = 1/(e^x + (1+e^(2x))^(1/2))#

#(du)/(dx) = e^x + d/(dx)((1+e^(2x))^(1/2))#

For the square root use chain rule again with

#phi = (1+e^(2x))^(1/2)#

#v(x) = 1 + e^(2x) and phi = v^(1/2)#

#(dv)/(dx) = 2e^(2x) and (dphi)/(dv) = 1/(2sqrt(v))#

#(dphi)/(dx) = (dphi)/(dv)(dv)/(dx) = (e^(2x))/(sqrt(1 + e^(2x)))#

#therefore (du)/(dx) = e^x + (e^(2x))/(sqrt(1 + e^(2x)))#

#(dy)/(dx) = (dy)/(du)(du)/(dx)#

#=1/(e^x + (1+e^(2x))^(1/2))*(e^x + (e^(2x))/(sqrt(1 + e^(2x))))#

#=e^x/(e^x + sqrt(1+e^(2x))) + e^(2x)/(sqrt(1 + e^(2x))(e^x+sqrt(1+e^(2x)))#

Bringing together over LCD:

#=(e^xsqrt(1+e^(2x)) + e^(2x))/(sqrt(1 + e^(2x))(e^x+sqrt(1+e^(2x)))#

Take factor of #e^x# out of numerator:

#=(e^x(sqrt(1+e^(2x)) + e^x))/(sqrt(1 + e^(2x))(e^x+sqrt(1+e^(2x)))#

Cancel out and obtain

#=(e^x)/(sqrt(1+e^(2x)))#