Linearization of a function using sqrt ln, confused with the denominator

Question

Can someone please explain why the answer to this question is $\frac{9}{7\cdot169}$ rather than $\frac{9}{7\cdot\sqrt[]{169}}$?

I understand this may seem simple, but the concept is not at all sinking into my head.

thank you for the help!!

ANYBODY!?!?!?!?!?!

Answer 1

We shall proceed with the very definition o the directional derivative, that is the change in the value of the function by a small change in the direction of the vector, divided by the norm of the vector. As such, we have:$$\begin{array}{c}f(x,y,z) = \ln \sqrt {{x^2} + {y^2} + {z^2}} \\{\rm{Answer}} = \mathop {\lim }\limits_{h \to 0} \frac{{f(3 + 3h,4 + 6h,12 - 2h) - f(3,4,12)}}{{\sqrt {{{\left( {3h} \right)}^2} + {{\left( {6h} \right)}^2} + {{\left( { - 2h} \right)}^2}} }}\\ = \mathop {\lim }\limits_{h \to 0} \frac{{\ln \sqrt {{{\left( {3 + 3h} \right)}^2} + {{\left( {4 + 6h} \right)}^2} + {{\left( {12 - 2h} \right)}^2}} - \ln \sqrt {{3^2} + {4^2} + {{12}^2}} }}{{7h}}\\ = \mathop {\lim }\limits_{h \to 0} \frac{{\ln \left( {\frac{{169 + 18h}}{{169}}} \right)}}{{14h}}\\ = \mathop {\lim }\limits_{h \to 0} \frac{{\ln \left( {1 + \frac{{18h}}{{169}}} \right)}}{{14h}}\\ = \mathop {\lim }\limits_{h \to 0} \frac{1}{{14h}}\frac{{18h}}{{169}} = \frac{9}{{1183}}\end{array} % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfKttLearuqr1ngBPrgarmqr1ngBPrgitL % xBI9gBamXvP5wqSXMqHnxAJn0BKvguHDwzZbqegm0B1jxALjhiov2D % aeHbuLwBLnhiov2DGi1BTfMBaebbfv3ySLgzGueE0jxyaibaieYlf9 % irVeeu0dXdh9vqqj-hEeeu0xXdbba9frFj0-OqFfea0dXdd9vqaq-J % frVkFHe9pgea0dXdar-Jb9hs0dXdbPYxe9vr0-vr0-vqpWqaaeaabi % GaciaacaqabeaadaabauaaaOabaqqabaGaemOzayMaeiikaGIaemiE % aGNaeiilaWIaemyEaKNaeiilaWIaemOEaONaeiykaKIaeyypa0Jagi % iBaWMaeiOBa42aaOaaaeaacqWG4baEdaahaaWcbeqaaiabikdaYaaa % kiabgUcaRiabdMha5naaCaaaleqabaGaeGOmaidaaOGaey4kaSIaem % OEaO3aaWbaaSqabeaacqaIYaGmaaaabeaaaOqaaiabbgeabjabb6ga % UjabbohaZjabbEha3jabbwgaLjabbkhaYjabg2da9maaxababaGagi % iBaWMaeiyAaKMaeiyBa0galeaacqWGObaAcqGHsgIRcqaIWaamaeqa % aOWaaSaaaeaacqWGMbGzcqGGOaakcqaIZaWmcqGHRaWkcqaIZaWmcq % WGObaAcqGGSaalcqaI0aancqGHRaWkcqaI2aGncqWGObaAcqGGSaal % cqaIXaqmcqaIYaGmcqGHsislcqaIYaGmcqWGObaAcqGGPaqkcqGHsi % slcqWGMbGzcqGGOaakcqaIZaWmcqGGSaalcqaI0aancqGGSaalcqaI % XaqmcqaIYaGmcqGGPaqkaeaadaGcaaqaamaabmaabaGaeG4mamJaem % iAaGgacaGLOaGaayzkaaWaaWbaaSqabeaacqaIYaGmaaGccqGHRaWk % daqadaqaaiabiAda2iabdIgaObGaayjkaiaawMcaamaaCaaaleqaba % GaeGOmaidaaOGaey4kaSYaaeWaaeaacqGHsislcqaIYaGmcqWGObaA % aiaawIcacaGLPaaadaahaaWcbeqaaiabikdaYaaaaeqaaaaaaOqaai % abg2da9maaxababaGagiiBaWMaeiyAaKMaeiyBa0galeaacqWGObaA % cqGHsgIRcqaIWaamaeqaaOWaaSaaaeaacyGGSbaBcqGGUbGBdaGcaa % qaamaabmaabaGaeG4mamJaey4kaSIaeG4mamJaemiAaGgacaGLOaGa % ayzkaaWaaWbaaSqabeaacqaIYaGmaaGccqGHRaWkdaqadaqaaiabis % da0iabgUcaRiabiAda2iabdIgaObGaayjkaiaawMcaamaaCaaaleqa % baGaeGOmaidaaOGaey4kaSYaaeWaaeaacqaIXaqmcqaIYaGmcqGHsi % slcqaIYaGmcqWGObaAaiaawIcacaGLPaaadaahaaWcbeqaaiabikda % YaaaaeqaaOGaeyOeI0IagiiBaWMaeiOBa42aaOaaaeaacqaIZaWmda % ahaaWcbeqaaiabikdaYaaakiabgUcaRiabisda0maaCaaaleqabaGa % eGOmaidaaOGaey4kaSIaeGymaeJaeGOmaiZaaWbaaSqabeaacqaIYa % GmaaaabeaaaOqaaiabiEda3iabdIgaObaaaeaacqGH9aqpdaWfqaqa % aiGbcYgaSjabcMgaPjabc2gaTbWcbaGaemiAaGMaeyOKH4QaeGimaa % dabeaakmaalaaabaGagiiBaWMaeiOBa42aaeWaaeaadaWcaaqaaiab % igdaXiabiAda2iabiMda5iabgUcaRiabigdaXiabiIda4iabdIgaOb % qaaiabigdaXiabiAda2iabiMda5aaaaiaawIcacaGLPaaaaeaacqaI % XaqmcqaI0aancqWGObaAaaaabaGaeyypa0ZaaCbeaeaacyGGSbaBcq % GGPbqAcqGGTbqBaSqaaiabdIgaOjabgkziUkabicdaWaqabaGcdaWc % aaqaaiGbcYgaSjabc6gaUnaabmaabaGaeGymaeJaey4kaSYaaSaaae % aacqaIXaqmcqaI4aaocqWGObaAaeaacqaIXaqmcqaI2aGncqaI5aqo % aaaacaGLOaGaayzkaaaabaGaeGymaeJaeGinaqJaemiAaGgaaaqaai % abg2da9maaxababaGagiiBaWMaeiyAaKMaeiyBa0galeaacqWGObaA % cqGHsgIRcqaIWaamaeqaaOWaaSaaaeaacqaIXaqmaeaacqaIXaqmcq % aI0aancqWGObaAaaWaaSaaaeaacqaIXaqmcqaI4aaocqWGObaAaeaa % cqaIXaqmcqaI2aGncqaI5aqoaaGaeyypa0ZaaSaaaeaacqaI5aqoae % aacqaIXaqmcqaIXaqmcqaI4aaocqaIZaWmaaaaaaa!1672! $$