Matematik haster:(

Benyt, at

y = cosh(x) = (e^x + e^-x)/2 , dvs

(e^x)² - 2y·e^x +1 = 0 , hvoraf

e^x = y ±√(y² -1) , hvor man benytter + , så

cosh^-1(x) = ln(x + √(x²-1)) , x ≥ 1 .

Heraf fås

^d/_dx(cosh^-1(x)) = (1 + x/√(x²-1))/(x + √(x²-1)) = 1/√(x²-1) .

man  hvad er forholdene for cos^-(x)?

#2

Benyt formlen for differentiation af invers funktion:

^d/_dx(cos^-1(x)) = 1/cos'(cos^-1(x)) = -1/sin(cos^-1(x)) = -1/√(1 - cos²(cos^-1(x))) = -1/√(1 - x²) , |x| < 1 .

tusind tak for hjælpen Andersen

#0  og  #1
        eller noteret

                               y = cosh^-1(x)        x ≥ 1

                               cosh(y) = x                                         som differentieres mht x

                               d(cosh(y))/dx = sinh(y)•(dy/dx) = 1
                       
                                                  1                    1                    1              
                               (dy/dx) = ---------- = ------------------- = ----------    x > 1
                                               sinh(y)     √(cosh²(y) -1)     √(x²-1)