Calculus of a Single Variable, Chapter 5, 5.8, Section 5.8, Problem 28

y=ln(tanh(x/2))
The derivative formula of natural logarithm is

d/dx[ln(u)] = 1/u*(du)/dx
Applying this formula, the derivative of the function will be
y' = d/dx [ln(tanh(x/2))]
y' = 1/(tanh(x/2)) * d/dx[tanh(x/2)]
To take the derivative of hyperbolic tangent, apply the formula

d/dx[tanh(u)] = sec h^2 (u) * (du)/dx
So y' will become
y'= 1/(tanh(x/2)) * sec h^2 (x/2) * d/dx(x/2)
y' = 1/(tanh(x/2)) *sec h^2(x/2) * 1/2
y'=(sec h^2(x/2))/(2tanh(x/2))
To simplify it further, express it in terms of hyperbolic sine and hyperbolic cosine.

sec h(u) = 1/cosh(u)

tanh(u)=sinh(u)/cosh(u)
Applying this, y' will become
y'= (1/(cosh^2(x/2)))/(2*sinh(x/2)/cosh(x/2))
y'= (1/(cosh^2(x/2)))/((2sinh(x/2))/cosh(x/2))
y'=1/(cosh^2(x/2)) * cosh(x/2)/(2sinh(x/2))
y'=1/cosh(x/2) * 1/(2sinh(x/2))
y'=1/(2sinh(x/2)cosh(x/2))
Then, apply the identity

sinh(2u) = 2sinh(u)cos(u)
So y' will be
y' = 1/sinh(2*x/2)
y'=1/sinh(x)

Therefore, the derivative of the given function is y'=1/sinh(x) .