By C. Plumpton, W. A. Tomkys

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INVERSE CIRCULAR FUNCTIONS 59 FIG. 106 (i). The graph of y = sinh"1^. y FIG. 106 (ii). The graph of y = cosh"1*. 6 Derivatives of the inverse hyperbolic functions If y = sinh -1 ^ so that x = sinhy, then dy 1 1 2 coshy V(sinh 2/ + 1) dx 1 j/(a: + 1) 2 the positive sign being taken for the radical because coshy is positive for all values of y. •••-^(-h-,) = W T Î Î . 24) 60 PURE MATHEMATICS ——(Cosh-1 a;) = ——— Similarly — where ~T~ y \X Q. X A) Cosh -1 a; is the two valued function. If we define y = cosh -1 a; as y = Cosh -1 a; for y ^ 0, then ^(cosh-1,) = - y F i _ .

41. # + Ci OtX* + b2x + c2 and ax 2 4- bx 4- c « 0 are equal. ) δχ Ci = 0. a2 62 c2 42. ) 40 PURE MATHEMATICS 43. (i) Show t h a t (x -f y -f- z) is a factor of the determinant ! y -f z x2 \z + x y2 \ x + y z2 x\ y\, z\ and find the remaining factors. (ii) If 2s = a + b + c, prove t h a t I sin(s — a) sina 11 sin (s — b) sin b 11 I sin(s — c) sine 1 | is equal to 4 sins sin\{b — c) sin J(c — a) sin£(a — b). ) 44. The three distinct straight lines y = mx — m2 + m3, y = nx — n2 + w3, y = px — p2 + p 3 meet in a point.

Area of A PN = I —asinhtasinhtdt J a o h = ab Ji(coah2t- l)dt o 1 . ,Λ 1 1*1 — ab — sinh2£ — —t 2 o 4 = — a b sinh2i 1 ——abt x . the required area OP A — — a 6 s i n h 2 ^ — ί — a&sinji2i 1 ——abtA = \abtx. A similar result is obtained for the ^-negative part of this branch of the curve where tx < 0. Hence the required area = \ \abtx\. , a = b], the area is H « 2 ^ which is analogous to the sectorial area \α2θ for a circle. (v) I=fsechxdx=j es+ *-s· Putting ex = u, so that dujdx = ex, gives 2 du ■f u +l 2 = 2 tan" 1u + (7 = 2 tan" * (e*) + C.