Now
d P d z = ∂ P ∂ ξ ∂ ξ ∂ z + ∂ P ∂ η ∂ η ∂ z + ∂ P ∂ ζ ∂ ζ ∂ z = ( b − c ) ∂ P ∂ ξ + ( c − a ) ∂ P ∂ η + ( a − b ) ∂ P ∂ ζ = ( b − c z − a + c − a z − b + a − b z − c ) θ H ′ ( θ ) = − ( b − c ) ( c − a ) ( a − b ) ( z − a ) ( z − b ) ( z − c ) θ H ′ ( θ ) . {\displaystyle {\begin{array}{cl}{\frac {dP}{dz}}&={\frac {\partial P}{\partial \xi }}{\frac {\partial \xi }{\partial z}}+{\frac {\partial P}{\partial \eta }}{\frac {\partial \eta }{\partial z}}+{\frac {\partial P}{\partial \zeta }}{\frac {\partial \zeta }{\partial z}}=(b-c){\frac {\partial P}{\partial \xi }}+\left(c-a\right){\frac {\partial P}{\partial \eta }}+(a-b){\frac {\partial P}{\partial \zeta }}\\\\&=\left({\frac {b-c}{z-a}}+{\frac {c-a}{z-b}}+{\frac {a-b}{z-c}}\right)\theta H'(\theta )\\\\&=-{\frac {(b-c)(c-a)(a-b)}{(z-a)(z-b)(z-c)}}\theta H'(\theta ).\end{array}}}
Hence
Also, since
we have
this relation being obtained in the same way as the one above.
The last expression may be written
i.e.,
or
and
consequently,