<s>[INST] 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?
 [/INST] 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: 
$$\frac{1}{2}\cdot\frac{-2z}{a^2-z^2}-\frac{1}{2}\cdot\frac{2z}{a^2+z^2}=\frac{1}{2} \frac{-2z(a^2+z^2)-2z(a^2-z^2)}{(a^2-z^2)(a^2+z^2)}=\\=\frac{1}{2} \frac{-4a^2z}{(a^2-z^2)(a^2+z^2)}=\frac{-2a^2z}{(a^2-z^2)(a^2+z^2)}$$
</s>