$\eqalign{
 & y = {2^x} + {2^{ - x}} \cr 
 & \ln y = x\ln 2 - x\ln 2 \cr 
 & \ln y = 0 \cr 
 & {1 \over y}{{dy} \over {dx}} = 0 \cr 
 & {{dy} \over {dx}} = 0 \cr} $

I've checked the answer and I've got the differential wrong, What am I doing wrong? I assume it has something to do with the expression with the negative exponent? Am I not allowed to prefix the natural expression with "minus" x? Could someone explain please, thank you!