Calculus: Early Transcendentals, Chapter 3, 3.6, Section 3.6, Problem 20

H(z)=lnsqrt((a^2-z^2)/(a^2+z^2))
Differentiating both sides with respect to z, we get
H'(z)=(1/sqrt((a^2-z^2)/(a^2+z^2))) d/dzsqrt((a^2-z^2)/(a^2+z^2))
H'(z)=sqrt((a^2+z^2)/(a^2-z^2)) ((sqrt(a^2+z^2)d/dzsqrt(a^2-z^2)-sqrt(a^2-z^2)d/dzsqrt(a^2+z^2))/(a^2+z^2))
H'(z)=sqrt((a^2+z^2)/(a^2-z^2)) ((sqrt(a^2+z^2)(1/2)(a^2-z^2)^(-1/2)(-2z)-sqrt(a^2-z^2)(1/2)(a^2+z^2)^(-1/2)(2z))/(a^2+z^2))
H'(z)=sqrt((a^2+z^2)/(a^2-z^2))((-zsqrt((a^2+z^2)/(a^2-z^2))-zsqrt((a^2-z^2)/(a^2+z^2)))/(a^2+z^2))
H'(z)=(((-z(a^2+z^2))/(a^2-z^2))-z)/(a^2+z^2)
H'(z)=(-z(a^2+z^2)-z(a^2-z^2))/((a^2-z^2)(a^2+z^2))
H'(z)=(-z(a^2+z^2+a^2-z^2))/(a^4-z^4)
H'(z)=(-2a^2z)/(a^4-z^4)