From (17) we get:
∂h
∂E = 1
E τ⅛
δ
1 β (1 - δ) = E ⅛β
1 + β - 2βδ 1 + β -
and also:
∂h
∂β
„1 (1 - δ) (1 + β - 2βδ) - β (1 - δ) (1 - 2δ)
E 1—0
(1 + β - 2βδ)2
■ 1 + β — 2βδ — δ — βδ + 2βδ7 — β + 2βδ + βδ — 2βδ7
Ej 1 — δ --------------------------------------------------------------------------------------------
(1 + β — 2 ••■■■ 2
ι 1 — δ
E I—0 -----------------
(1+ β — 2βδ)2
and then:
∂h
∂δ
1 — β (1 + β — 2βδ) + 2β2 (1 — δ)
Ej 1—0 -------------------------------------------------------
(1 + β — 2βδ)2
β(1 — δ) 1 1
1 + β — 2βδE—S (1 — δ)2
log E =
E —β — β2 + 2β2δ + 2β2 — 2β2δ + Eʌ β l σ E
(1 + β — 2βδ)2 + (1 — δ)(1 + β — 2βδ)lθg
1
E1-0
β2 — β
(1+ β — 2βδ)2
1
β
(1 — δ)(1 + β — 2βδ)
log E =
E 1⅛ ∣^ β (β — 1) , β
1 + β — 2βδ [1 + β — 2βδ + 1 — δ
From (18) we have:
∂ρ = (α (ξ + η + θ) — ξ) η — αη (ξ + η + θ) =
dα ^ (α (ξ + η + θ) — ξ)2 ^
αξη + αη2 + αθη — ξη — αξη — αη2 — αθη ξη
(a (ξ + η + θ) — ξ)2 (a (ξ + η + θ) — ξ)2
and then:
∂ρ -№п ∙ α α2η
dθ (α (ξ + η + θ) — ξ)2 (α (ξ + η + θ) — ξ)2
and also:
∂ρ —oq (α — 1) αη (1 — α)
dξ (a (ξ + η + θ) — ξ)2 (a (ξ + η + θ) — ξ)2
and finally:
∂ρ (α (ξ + η + θ) — ξ) α — αη ∙ α
dη ~ (α (ξ + η + θ) — ξ)2 "^
α2ξ + α2η + α2θ — αξ — α2η α [α (ξ + θ) — ξ]
(a (ξ + η + θ) — ξ)2 (a (ξ + η + θ) — ξ)2
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