Calculus of a Single Variable, Chapter 5, 5.8, Section 5.8, Problem 31

f(t)=arctan(sinh(t))
Take note that the derivative formula of arctangent is
d/dx[arctan(u)]=1/(1+u^2)*(du)/dx
Applying this, the derivative of the function will be
f'(t) = d/(dt)[arctan(sinh(t))]
f'(t) = 1/(1+sinh^2(t)) *d/(dt)[sinh(t)]
Also, the derivative formula of hyperbolic sine is
d/dx[sinh(u)]=cosh(u)*(du)/(dx)
Applying this, f'(t) will become
f'(t)= 1/(1+sinh^2(t)) *cosh(t)*d/(dt)(t)
f'(t)= 1/(1+sinh^2(t)) *cosh(t)*1
f'(t)= cosh(t)/(1+sinh^2(t))
f'(t)= cosh(t)/(cosh^2(t))
f'(t)= 1/cosh(t) is the final derivative