Single Variable Calculus, Chapter 7, 7.4-1, Section 7.4-1, Problem 20

Determine the function $\displaystyle H(z) = \ln \sqrt{\frac{a^2 - z^2}{a^2 + z^2}}$


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\begin{equation}
\begin{aligned}
H'(z) &= \frac{1}{\sqrt{\frac{a^2-z^2}{a^2+z^2}}} \cdot \frac{d}{dz} \sqrt{\frac{a^2-z^2}{a^2+z^2}}\\
\\
H'(z) &= \frac{1}{\frac{\left( a^2 - z^2\right)}{\left( a^2 + z^2\right)}} \cdot \frac{d}{dz} \left( \frac{a^2 - z^2}{a^2 + z^2} \right)^{\frac{1}{2}}\\
\\
H'(z) &= \frac{\left(a^2 + z^2 \right)^{\frac{1}{2}}}{\left( a^2 - z^2 \right)^{\frac{1}{2}}} \cdot \frac{1}{2} \left( \frac{a^2-z^2}{a^2 + z^2} \right)^{\frac{-1}{2}} \frac{d}{dz} \left( \frac{a^2 - z^2}{a^2 + z^2} \right)\\
\\
H'(z) &= \frac{\left( a^z + z^2\right)^{\frac{1}{2}}}{\left( a^2 - z^2 \right)^{\frac{1}{2}}} \cdot \frac{1}{2} \frac{\left( a^2 - z^2 \right)^{\frac{-1}{2}}}{\left( a^2 + z^2 \right)^{\frac{-1}{2}}} \left[ \frac{\left(a^2 +z^2 \right)\frac{d}{dz}\left(a^2 - z^2 \right)-\left( a^2 - z^2 \right)\frac{d}{dz}\left( a^2 + z^2 \right) }{\left( a^2 + z^2 \right)^2} \right]\\
\\
H'(z) &= \frac{\left( a^2 + z^2 \right)^{\frac{1}{2}}}{\left( a^2 - z^2 \right)^{\frac{1}{2}}} \cdot \frac{\left( a^2 + z^2 \right)^{\frac{1}{2}}}{2\left( a^2 - z^2\right)^{\frac{1}{2}}} \left[ \frac{\left( a^2 + z^2 \right)(-2z) - \left( a^2 - z^2 \right)(2z)}{\left( a^2 + z^2 \right)^2} \right]\\
\\
H'(z) &= \frac{a^2 + z^2}{2(a^2 - z^2)} \left[ \frac{2z\left( -a^2 - \cancel{z^2} - a^2 + \cancel{z^2} \right)}{\left( a^2 + z^2 \right)^2} \right]\\
\\
H'(z) &= \frac{a^2 + z^2}{\cancel{2}\left( a^2 - z^2 \right)} \left[ \frac{\cancel{2}z(-2a^2)}{\left( a^2 + z^2 \right)^2} \right]\\
\\
H'(z) &= \frac{-2a^2 z}{\left( a^2 - z^2 \right)\left( a^2 + z^2 \right)}\\
\\
H'(z) &= \frac{-2a^2z}{a^4 - z^4}
\end{aligned}
\end{equation}
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