differentiate the function.\n\n$h(z)=lnleft(sqrt{\frac{a^{2}-z^{2}}{a^{2}+z^{2}}}\right)$\n\n$h(z)=$

Explanation:

Step1: Simplify the function

Use the logarithm property (\ln\sqrt{x}=\frac{1}{2}\ln x). (H(z)=\frac{1}{2}\ln\left(\frac{a^{2}-z^{2}}{a^{2}+z^{2}}\right)=\frac{1}{2}\left[\ln(a^{2}-z^{2})-\ln(a^{2}+z^{2})\right])

Step2: Differentiate using the chain rule

The chain rule is (\frac{d}{dz}[\ln(u)]=\frac{u'}{u}). For (y = \frac{1}{2}\ln(a^{2}-z^{2})), let (u=a^{2}-z^{2}), then (y'=\frac{1}{2}\cdot\frac{-2z}{a^{2}-z^{2}}). For (y =-\frac{1}{2}\ln(a^{2}+z^{2})), let (u=a^{2}+z^{2}), then (y'=-\frac{1}{2}\cdot\frac{2z}{a^{2}+z^{2}}).

Step3: Combine the derivatives

(H'(z)=\frac{1}{2}\left(\frac{-2z}{a^{2}-z^{2}}-\frac{2z}{a^{2}+z^{2}}\right)) Simplify the expression: [ \begin{align*} H'(z)&=\frac{-z(a^{2}+z^{2})-z(a^{2}-z^{2})}{(a^{2}-z^{2})(a^{2}+z^{2})}\ &=\frac{-za^{2}-z^{3}-za^{2}+z^{3}}{a^{4}-z^{4}}\ &=\frac{-2za^{2}}{a^{4}-z^{4}} \end{align*} ]

Answer:

(\frac{-2za^{2}}{a^{4}-z^{4}})