I only need a detailed breakdown of how they got dh/dxh2o (shown in picture). Please show step by step.

LOUBIANA STATE UNIVERSITY Exercise dk k = K1 + x2 dx2 Develop an expression for the partial molar enthalpy of sulfuric acid in water at 21C using Equation dh HH2504 = h XH20 dx H20 - h = 1.596XH,so, + 1.591x4,0 74.40x4,80X2,0 (1 0.56LxH.so,)[kJ/mol] -1.596 + 1.591 - 74.40[(xH:so, - 10) (1 - 0.56lxn.so.) + 0.561211,0 :O] - dh dxo TY