how differtial wokrs in quantom calculas firts result of quantom calculas new calculas

n0 takens because its free charge and wher to go it undetermined imagine qx = q and qy = qxcube then in such manners thers is f = q1.q2/rsqure so how dqy/dqx is psoisibel dqxcube/dqx df =dqxcube*dqx/drsquqre so df/dqx=dqxcube/drsquare dqxcube/dqxdf = (dqxcube)square/drsquare for left side (dqxcube)squqre thre will be neutral sigantuer will come (dqxcube/dqx).df = (dqxcube)square/drsquare (dqxcube/dqx).df = n0/drsquare then df = n0/drsquqre(dqx/dqxcube) df =(dqxcube/dqx).n0 then normal intergar parts f = (qxpow fore+1/four+1/qx).n0

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