Differentiate the Function $ h(z)=\ln\sqrt{\frac{a^2-z^2}{a^2+z^2}}$ 
Differentiate the function $$h(z)=\ln\sqrt{\frac{a^2-z^2}{a^2+z^2}}$$

My try:
$$h(z) = \frac{1}{2}\ln\left(a^2-z^2\right)-\frac{1}{2}\ln\left(a^2+z^2\right)$$
so
$$h'(z) = \frac{1}{2}\cdot\frac{2a-2z}{a^2-z^2}-\frac{1}{2}\cdot\frac{2a+2z}{a^2+z^2}$$
My answer is therefore $$h'(z) = \frac{-2a^2z+2z^2a}{(a^2-z^2)(a^2+z^2)}$$
Is it correct?
 A: No, this is not correct, though you are on the good path.  $a^2$ is a constant, so when differentiated, it will be gone. The correct derivative is: 
$$\frac{1}{2}\cdot\frac{-2z}{a^2-z^2}-\frac{1}{2}\cdot\frac{2z}{a^2+z^2}=\frac{1}{2} \frac{-2z(a^2+z^2)-2z(a^2-z^2)}{(a^2-z^2)(a^2+z^2)}=\\=\frac{1}{2} \frac{-4a^2z}{(a^2-z^2)(a^2+z^2)}=\frac{-2a^2z}{(a^2-z^2)(a^2+z^2)}$$
A: Albeit you used logarithmic differentiation, you must never forget the power of substitutions.  
Let $$u=\sqrt{\frac{a^2-z^2}{a^2+z^2}}.$$ And, for more simplicity, let $$v=\frac{a^2-z^2}{a^2+z^2}. $$ Then $u=\sqrt{v} $ with $h(z)=\ln(u).$ 
We find: $${d}h(z)=\frac{du}{u}$$ $$du=\frac{1}{2}v^{-1/2}dv $$ $$dv=\frac{-2z}{a^2+z^2}-\frac{2z(a^2-z^2)}{(a^2+z^2)^2}dz. $$
Before we undo our substitutions, we mentally note that $$\frac{dh(z)}{du}\times\frac{du}{dv}\times\frac{dv}{dz}=\frac{d}{dz}h(z)=h'(z). $$
Substituting back in yields the final result of $$h'(z)=\frac{-z}{a^2-z^2}-\frac{z}{a^2+z^2}. $$ To match other results on this page, a little algebraic manipulation yields the result $$h'(z)=\frac{-2a^2z}{(a^2-z^2)(a^2+z^2)}. $$
