$$-(\nu+1)\frac{d}{d\nu}\left[\ln\Gamma\left(\frac{\nu}{2}\right)-\ln\Gamma\left(\frac{\nu+1}{2}\right)\right]=\frac{\nu+1}{2}\left[\psi\left(\frac{\nu+1}{2}\right)-\psi\left(\frac{\nu}{2}\right)\right]$$
How to derive the above equality? I don't understand how taking the derivative of the log gamma function results in the digamma functions and other resulting side-effects. What steps occurred in between?
For the sake of answering the question we have from the definition of the polygamma function and the chain rule: $$ (\nu+1)\partial_\nu^{n+1}(\log\Gamma(\tfrac{\nu+1}{2})-\log\Gamma(\tfrac{\nu}{2}))=(\nu+1)2^{-n-1}(\psi^{(n)}(\tfrac{\nu+1}{2})-\psi^{(n)}(\tfrac{\nu}{2})), $$ for which your problem is given by the special case $n=0$.