YoungKukie24
03.11.2020 •
Mathematics
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
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