Prove that d dx (sinh−1(x)) = 1 1 + x2 . Solution 1 Let y = sinh−1(x). Then sinh(y) = x. If we differentiate this equation implicitly with respect to x, we get dy dx = 1. Since cosh2(y) − sinh2(y) = 1 and cosh(y) ≥ 0, we have cosh(y) = 1 + sinh2(y) , so dy dx = 1 cosh(y) = 1 1 + sinh2(y) = . Solution 2 From the equation sinh−1(x) = ln  x + x2 + 1 , we have d dx  (sinh−1(x)) = d dx  ln x + x2 + 1 = 1 x + x2 + 1   d dx   x + x2 + 1 = 1 x + x2+ 1   1 + x x2 + 1 = x2 + 1 + x x + x2 + 1   x2 + 1 = 1 x2 + 1 g