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

1 Answer
Aug 9, 2016

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

Explanation:

Use the chain rule.

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)))