with respect to x  y=sin3x y+δy=sin3(x+δx) δy=sin3(x+δx)−sin3x δy=sin(3x+3δx)−sin3x δy=2cos 3x+3δx+3x 2 sin 3x+3δx−3x 2 | note:sinA−sinB=2cos A+B 2 sin A−B 2 δy=2cos(3x+ 3δy 2 )sin 3δx 2 Divide through by δx δy δx = 2cos(3x+ 3δx 2 )sin 3δx 2 δx = 2cos(3x+ 3δx 2 ) 1 ⋅ sin 3δx 2 δx δy δx = cos(3x+ 3δx 2 ) 1 sin 3δx 2 ×3 3δx 2 ×3 Note the step we have it in the form  sinθ θ Proceeding to the limits as δx→0, we get dy dx = lim δx→0 δy δx =3⋅ lim δx→0 cos(3x+ 3δx 2 ) 1 ⋅ lim δx→0 sin 3δx 2 3δx 2 | Note: lim δx→0 sin 3δx 2 3δx 2 =1 dy dx =3×cos3x×1=3cos3x Hence,  d dx (sin3x)=3cos3x MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakqaabeqaaiaadMhacq GH9aqpciGGZbGaaiyAaiaac6gacaaIZaGaamiEaaqaaiaadMhacqGH RaWkcqaH0oazcaWG5bGaeyypa0Jaci4CaiaacMgacaGGUbGaaG4mai 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iEaiabgUcaRmaaleaaleaacaaIZaGaeqiTdqMaamiEaaqaaiaaikda aaGccaGGPaaabaGaaGymaaaacqGHflY1daWfqaqaaiGacYgacaGGPb GaaiyBaaWcbaGaeqiTdqMaamiEaiabgkziUkaaicdaaeqaaOWaaSaa aeaaciGGZbGaaiyAaiaac6gadaWcbaWcbaGaaG4maiabes7aKjaadI haaeaacaaIYaaaaaGcbaWaaSqaaSqaaiaaiodacqaH0oazcaWG4baa baGaaGOmaaaaaaGcdaabbaqaaiaad6eacaWGVbGaamiDaiaadwgaca GG6aWaaCbeaeaaciGGSbGaaiyAaiaac2gaaSqaaiabes7aKjaadIha cqGHsgIRcaaIWaaabeaakmaalaaabaGaci4CaiaacMgacaGGUbWaaS qaaSqaaiaaiodacqaH0oazcaWG4baabaGaaGOmaaaaaOqaamaaleaa leaacaaIZaGaeqiTdqMaamiEaaqaaiaaikdaaaaaaaGccaGLhWoacq GH9aqpcaaIXaaabaWaaSaaaeaacaWGKbGaamyEaaqaaiaadsgacaWG 4baaaiabg2da9iaaiodacqGHxdaTciGGJbGaai4BaiaacohacaaIZa GaamiEaiabgEna0kaaigdacqGH9aqpcaaIZaGaci4yaiaac+gacaGG ZbGaaG4maiaadIhaaeaacaWGibGaamyzaiaad6gacaWGJbGaamyzai aacYcacaqGGaWaaSaaaeaacaWGKbaabaGaamizaiaadIhaaaGaaiik aiGacohacaGGPbGaaiOBaiaaiodacaWG4bGaaiykaiabg2da9iaaio daciGGJbGaai4BaiaacohacaaIZaGaamiEaaaaaa@FF05@ Determine from first principle the derivative of y= x MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWG5bGaeyypa0 ZaaOaaaeaacaWG4baaleqaaaaa@38A0@ with respect to x y= x y+δy= x+δx δy= x+δx − x δy= x+δx − x × x+δx + x x+δx + x δy= x+δx−x x+δx + x = δx x+δx + x Dividing through by δx δy δx = δx δx[ x+δx + x ] = 1 x+δx + x Proceed to the limitsas δx→0 dy dx = lim δx→0 δy δx = 1 x+0 + x = 1 2 x Hence  d dx ( x )= 1 2 x MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakqaabeqaaiaadMhacq GH9aqpdaGcaaqaaiaadIhaaSqabaaakeaacaWG5bGaey4kaSIaeqiT dqMaamyEaiabg2da9maakaaabaGaamiEaiabgUcaRiabes7aKjaadI haaSqabaaakeaacqaH0oazcaWG5bGaeyypa0ZaaOaaaeaacaWG4bGa ey4kaSIaeqiTdqMaamiEaaWcbeaakiabgkHiTmaakaaabaGaamiEaa WcbeaaaOqaaiabes7aKjaadMhacqGH9aqpdaGcaaqaaiaadIhacqGH RaWkcqaH0oazcaWG4baaleqaaOGaeyOeI0YaaOaaaeaacaWG4baale qaaOGaey41aq7aaSaaaeaadaGcaaqaaiaadIhacqGHRaWkcqaH0oaz caWG4baaleqaaOGaey4kaSYaaOaaaeaacaWG4baaleqaaaGcbaWaaO aaaeaacaWG4bGaey4kaSIaeqiTdqMaamiEaaWcbeaakiabgUcaRmaa kaaabaGaamiEaaWcbeaaaaaakeaacqaH0oazcaWG5bGaeyypa0ZaaS aaaeaacaWG4bGaey4kaSIaeqiTdqMaamiEaiabgkHiTiaadIhaaeaa daGcaaqaaiaadIhacqGHRaWkcqaH0oazcaWG4baaleqaaOGaey4kaS YaaOaaaeaacaWG4baaleqaaaaakiabg2da9maalaaabaGaeqiTdqMa amiEaaqaamaakaaabaGaamiEaiabgUcaRiabes7aKjaadIhaaSqaba GccqGHRaWkdaGcaaqaaiaadIhaaSqabaaaaaGcbaGaaeiraiaabMga caqG2bGaaeyAaiaabsgacaqGPbGaaeOBaiaabEgacaqGGaGaaeiDai aabIgacaqGYbGaae4BaiaabwhacaqGNbGaaeiAaiaabccacaqGIbGa aeyEaiaabccacqaH0oazcaWG4baabaWaaSaaaeaacqaH0oazcaWG5b aabaGaeqiTdqMaamiEaaaacqGH9aqpdaWcaaqaaiabes7aKjaadIha aeaacqaH0oazcaWG4bGaai4wamaakaaabaGaamiEaiabgUcaRiabes 7aKjaadIhaaSqabaGccqGHRaWkdaGcaaqaaiaadIhaaSqabaGccaGG Dbaaaiabg2da9maalaaabaGaaGymaaqaamaakaaabaGaamiEaiabgU caRiabes7aKjaadIhaaSqabaGccqGHRaWkdaGcaaqaaiaadIhaaSqa baaaaaGcbaGaaeiuaiaabkhacaqGVbGaae4yaiaabwgacaqGLbGaae izaiaabccacaqG0bGaae4BaiaabccacaqG0bGaaeiAaiaabwgacaqG GaGaaeiBaiaabMgacaqGTbGaaeyAaiaabshacaqGZbGaaeyyaiaabo hacaqGGaGaeqiTdqMaamiEaiabgkziUkaaicdaaeaadaWcaaqaaiaa dsgacaWG5baabaGaamizaiaadIhaaaGaeyypa0ZaaCbeaeaaciGGSb GaaiyAaiaac2gaaSqaaiabes7aKjaadIhacqGHsgIRcaaIWaaabeaa kmaalaaabaGaeqiTdqMaamyEaaqaaiabes7aKjaadIhaaaGaeyypa0 ZaaSaaaeaacaaIXaaabaWaaOaaaeaacaWG4bGaey4kaSIaaGimaaWc beaakiabgUcaRmaakaaabaGaamiEaaWcbeaaaaGccqGH9aqpdaWcaa qaaiaaigdaaeaacaaIYaWaaOaaaeaacaWG4baaleqaaaaaaOqaaiaa bIeacaqGLbGaaeOBaiaabogacaqGLbGaaeiiamaalaaabaGaamizaa qaaiaadsgacaWG4baaamaabmaabaWaaOaaaeaacaWG4baaleqaaaGc caGLOaGaayzkaaGaeyypa0ZaaSaaaeaacaaIXaaabaGaaGOmamaaka aabaGaamiEaaWcbeaaaaaaaaa@F8A2@  Determine from first principle the derivative of logx MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaciGGSbGaai4Bai aacEgacaWG4baaaa@3951@ with respect to x y= log e x y+δy= log e (x+δx) δy= log e (x+δx)− log e x δy= log e ( x+δx x )= log e ( 1+ δx x ) Dividing through by δx δy δx = 1 δx log e ( 1+ δx x )= 1 x × x δx log( 1+ δx x ) δy δx = 1 x log e ( 1+ δx x ) x δx = 1 x log e ( 1+ 1 x δx ) x δx Put  x δx =p δy δx = 1 x log ( 1+ 1 p ) p Proceeding the their limitδx→0, p→∞,  ( 1+ 1 p ) p →e dy dx = lim δx→0 1 x log e ( 1+ 1 p ) p = 1 x log e e dy dx = 1 x Hence  d dx ( log e x)= 1 x MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakqaabeqaaiaadMhacq GH9aqpciGGSbGaai4BaiaacEgadaWgaaWcbaGaamyzaaqabaGccaWG 4baabaGaamyEaiabgUcaRiabes7aKjaadMhacqGH9aqpciGGSbGaai 4BaiaacEgadaWgaaWcbaGaamyzaaqabaGccaGGOaGaamiEaiabgUca Riabes7aKjaadIhacaGGPaaabaGaeqiTdqMaamyEaiabg2da9iGacY gacaGGVbGaai4zamaaBaaaleaacaWGLbaabeaakiaacIcacaWG4bGa ey4kaSIaeqiTdqMaamiEaiaacMcacqGHsislciGGSbGaai4BaiaacE gadaWgaaWcbaGaamyzaaqabaGccaWG4baabaGaeqiTdqMaamyEaiab g2da9iGacYgacaGGVbGaai4zamaaBaaaleaacaWGLbaabeaakmaabm aabaWaaSaaaeaacaWG4bGaey4kaSIaeqiTdqMaamiEaaqaaiaadIha aaaacaGLOaGaayzkaaGaeyypa0JaciiBaiaac+gacaGGNbWaaSbaaS qaaiaadwgaaeqaaOWaaeWaaeaacaaIXaGaey4kaSYaaSaaaeaacqaH 0oazcaWG4baabaGaamiEaaaaaiaawIcacaGLPaaaaeaacaqGebGaae yAaiaabAhacaqGPbGaaeizaiaabMgacaqGUbGaae4zaiaabccacaqG 0bGaaeiAaiaabkhacaqGVbGaaeyDaiaabEgacaqGObGaaeiiaiaabk gacaqG5bGaaeiiaiabes7aKjaadIhaaeaadaWcaaqaaiabes7aKjaa dMhaaeaacqaH0oazcaWG4baaaiabg2da9maalaaabaGaaGymaaqaai abes7aKjaadIhaaaGaciiBaiaac+gacaGGNbWaaSbaaSqaaiaadwga aeqaaOWaaeWaaeaacaaIXaGaey4kaSYaaSaaaeaacqaH0oazcaWG4b aabaGaamiEaaaaaiaawIcacaGLPaaacqGH9aqpdaWcaaqaaiaaigda aeaacaWG4baaaiabgEna0oaalaaabaGaamiEaaqaaiabes7aKjaadI haaaGaciiBaiaac+gacaGGNbWaaeWaaeaacaaIXaGaey4kaSYaaSaa aeaacqaH0oazcaWG4baabaGaamiEaaaaaiaawIcacaGLPaaaaeaada Wcaaqaaiabes7aKjaadMhaaeaacqaH0oazcaWG4baaaiabg2da9maa laaabaGaaGymaaqaaiaadIhaaaGaciiBaiaac+gacaGGNbWaaSbaaS qaaiaadwgaaeqaaOWaaeWaaeaacaaIXaGaey4kaSYaaSaaaeaacqaH 0oazcaWG4baabaGaamiEaaaaaiaawIcacaGLPaaadaahaaWcbeqaam aaleaameaacaWG4baabaGaeqiTdqMaamiEaaaaaaGccqGH9aqpdaWc aaqaaiaaigdaaeaacaWG4baaaiGacYgacaGGVbGaai4zamaaBaaale aacaWGLbaabeaakmaabmaabaGaaGymaiabgUcaRmaalaaabaGaaGym aaqaamaaleaaleaacaWG4baabaGaeqiTdqMaamiEaaaaaaaakiaawI cacaGLPaaadaahaaWcbeqaamaaleaameaacaWG4baabaGaeqiTdqMa amiEaaaaaaaakeaacaqGqbGaaeyDaiaabshacaqGGaWaaSaaaeaaca WG4baabaGaeqiTdqMaamiEaaaacqGH9aqpcaWGWbaabaWaaSaaaeaa cqaH0oazcaWG5baabaGaeqiTdqMaamiEaaaacqGH9aqpdaWcaaqaai aaigdaaeaacaWG4baaaiGacYgacaGGVbGaai4zamaabmaabaGaaGym aiabgUcaRmaalaaabaGaaGymaaqaaiaadchaaaaacaGLOaGaayzkaa WaaWbaaSqabeaacaWGWbaaaaGcbaGaaeiuaiaabkhacaqGVbGaae4y aiaabwgacaqGLbGaaeizaiaabMgacaqGUbGaae4zaiaabccacaqG0b GaaeiAaiaabwgacaqGGaGaaeiDaiaabIgacaqGLbGaaeyAaiaabkha caqGGaGaaeiBaiaabMgacaqGTbGaaeyAaiaabshacqaH0oazcaWG4b GaeyOKH4QaaGimaiaacYcacaqGGaGaamiCaiabgkziUkabg6HiLkaa cYcacaqGGaWaaeWaaeaacaaIXaGaey4kaSYaaSaaaeaacaaIXaaaba GaamiCaaaaaiaawIcacaGLPaaadaahaaWcbeqaaiaadchaaaGccqGH sgIRcaWGLbaabaWaaSaaaeaacaWGKbGaamyEaaqaaiaadsgacaWG4b aaaiabg2da9maaxababaGaciiBaiaacMgacaGGTbaaleaacqaH0oaz caWG4bGaeyOKH4QaaGimaaqabaGcdaWcaaqaaiaaigdaaeaacaWG4b aaaiGacYgacaGGVbGaai4zamaaBaaaleaacaWGLbaabeaakmaabmaa baGaaGymaiabgUcaRmaalaaabaGaaGymaaqaaiaadchaaaaacaGLOa GaayzkaaWaaWbaaSqabeaacaWGWbaaaOGaeyypa0ZaaSaaaeaacaaI XaaabaGaamiEaaaaciGGSbGaai4BaiaacEgadaWgaaWcbaGaamyzaa qabaGccaWGLbaabaWaaSaaaeaacaWGKbGaamyEaaqaaiaadsgacaWG 4baaaiabg2da9maalaaabaGaaGymaaqaaiaadIhaaaaabaGaaeisai aabwgacaqGUbGaae4yaiaabwgacaqGGaWaaSaaaeaacaWGKbaabaGa amizaiaadIhaaaGaaiikaiGacYgacaGGVbGaai4zamaaBaaaleaaca WGLbaabeaakiaadIhacaGGPaGaeyypa0ZaaSaaaeaacaaIXaaabaGa amiEaaaaaaaa@5CBF@                  Determine from first principle the derivative of tanx MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaciGG0bGaaiyyai aac6gacaWG4baaaa@3952@ with respect to x y=tanx y= sinx cosx y+δy= sin(x+δx) cos(x+δx) δy= sin(x+δx) cos(x+δx) − sinx cosx δy= sin(x+δx)cosx−sinxcos(x+δx) cosx(cosx+δx) δy= sin[(x+δx)−x] cosxcos(x+δx) = sinδx cosxcos(x+δx) Dividing through δx δy δx = sinδx δx⋅cos(x+δx)cosx = 1 cos(x+δx)cosx ⋅ sinδx δx Proceeding to to the limit as δx→0, we have dy dx = lim x→0 δy δx = 1 cos(x+0)cosx ×1 dy dx = 1 cos 2 x = sec 2 x Hence  d dx ( tanx )= sec 2 x MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakqaabeqaaiaadMhacq GH9aqpciGG0bGaaiyyaiaac6gacaWG4baabaGaamyEaiabg2da9maa laaabaGaci4CaiaacMgacaGGUbGaamiEaaqaaiGacogacaGGVbGaai 4CaiaadIhaaaaabaGaamyEaiabgUcaRiabes7aKjaadMhacqGH9aqp daWcaaqaaiGacohacaGGPbGaaiOBaiaacIcacaWG4bGaey4kaSIaeq iTdqMaamiEaiaacMcaaeaaciGGJbGaai4BaiaacohacaGGOaGaamiE aiabgUcaRiabes7aKjaadIhacaGGPaaaaaqaaiabes7aKjaadMhacq GH9aqpdaWcaaqaaiGacohacaGGPbGaaiOBaiaacIcacaWG4bGaey4k aSIaeqiTdqMaamiEaiaacMcaaeaaciGGJbGaai4BaiaacohacaGGOa GaamiEaiabgUcaRiabes7aKjaadIhacaGGPaaaaiabgkHiTmaalaaa baGaci4CaiaacMgacaGGUbGaamiEaaqaaiGacogacaGGVbGaai4Cai aadIhaaaaabaGaeqiTdqMaamyEaiabg2da9maalaaabaGaci4Caiaa cMgacaGGUbGaaiikaiaadIhacqGHRaWkcqaH0oazcaWG4bGaaiykai GacogacaGGVbGaai4CaiaadIhacqGHsislciGGZbGaaiyAaiaac6ga caWG4bGaci4yaiaac+gacaGGZbGaaiikaiaadIhacqGHRaWkcqaH0o azcaWG4bGaaiykaaqaaiGacogacaGGVbGaai4CaiaadIhacaGGOaGa ci4yaiaac+gacaGGZbGaamiEaiabgUcaRiabes7aKjaadIhacaGGPa aaaaqaaiabes7aKjaadMhacqGH9aqpdaWcaaqaaiGacohacaGGPbGa aiOBaiaacUfacaGGOaGaamiEaiabgUcaRiabes7aKjaadIhacaGGPa GaeyOeI0IaamiEaiaac2faaeaaciGGJbGaai4BaiaacohacaWG4bGa ci4yaiaac+gacaGGZbGaaiikaiaadIhacqGHRaWkcqaH0oazcaWG4b GaaiykaaaacqGH9aqpdaWcaaqaaiGacohacaGGPbGaaiOBaiabes7a KjaadIhaaeaaciGGJbGaai4BaiaacohacaWG4bGaci4yaiaac+gaca GGZbGaaiikaiaadIhacqGHRaWkcqaH0oazcaWG4bGaaiykaaaaaeaa caqGebGaaeyAaiaabAhacaqGPbGaaeizaiaabMgacaqGUbGaae4zai aabccacaqG0bGaaeiAaiaabkhacaqGVbGaaeyDaiaabEgacaqGObGa aeiiaiabes7aKjaadIhaaeaadaWcaaqaaiabes7aKjaadMhaaeaacq aH0oazcaWG4baaaiabg2da9maalaaabaGaci4CaiaacMgacaGGUbGa eqiTdqMaamiEaaqaaiabes7aKjaadIhacqGHflY1ciGGJbGaai4Bai aacohacaGGOaGaamiEaiabgUcaRiabes7aKjaadIhacaGGPaGaci4y aiaac+gacaGGZbGaamiEaaaacqGH9aqpdaWcaaqaaiaaigdaaeaaci GGJbGaai4BaiaacohacaGGOaGaamiEaiabgUcaRiabes7aKjaadIha caGGPaGaci4yaiaac+gacaGGZbGaamiEaaaacqGHflY1daWcaaqaai GacohacaGGPbGaaiOBaiabes7aKjaadIhaaeaacqaH0oazcaWG4baa aaqaaiaabcfacaqGYbGaae4BaiaabogacaqGLbGaaeyzaiaabsgaca qGPbGaaeOBaiaabEgacaqGGaGaaeiDaiaab+gacaqGGaGaaeiDaiaa b+gacaqGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaabYgacaqGPbGaae yBaiaabMgacaqG0bGaaeiiaiaabggacaqGZbGaaeiiaiabes7aKjaa dIhacqGHsgIRcaaIWaGaaiilaiaabccacaqG3bGaaeyzaiaabccaca qGObGaaeyyaiaabAhacaqGLbaabaWaaSaaaeaacaWGKbGaamyEaaqa aiaadsgacaWG4baaaiabg2da9maaxababaGaciiBaiaacMgacaGGTb aaleaacaWG4bGaeyOKH4QaaGimaaqabaGcdaWcaaqaaiabes7aKjaa dMhaaeaacqaH0oazcaWG4baaaiabg2da9maalaaabaGaaGymaaqaai GacogacaGGVbGaai4CaiaacIcacaWG4bGaey4kaSIaaGimaiaacMca ciGGJbGaai4BaiaacohacaWG4baaaiabgEna0kaaigdaaeaadaWcaa qaaiaadsgacaWG5baabaGaamizaiaadIhaaaGaeyypa0ZaaSaaaeaa caaIXaaabaGaci4yaiaac+gacaGGZbWaaWbaaSqabeaacaaIYaaaaO GaamiEaaaacqGH9aqpciGGZbGaaiyzaiaacogadaahaaWcbeqaaiaa ikdaaaGccaWG4baabaGaaeisaiaabwgacaqGUbGaae4yaiaabwgaca qGGaWaaSaaaeaacaWGKbaabaGaamizaiaadIhaaaWaaeWaaeaaciGG 0bGaaiyyaiaac6gacaWG4baacaGLOaGaayzkaaGaeyypa0Jaci4Cai aacwgacaGGJbWaaWbaaSqabeaacaaIYaaaaOGaamiEaaaaaa@85B4@  Determine from first principle the derivative of xsecx MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWG4bGaci4Cai aacwgacaGGJbGaamiEaaaa@3A47@ with respect to x y=xsecx y+δy=(x+δx)sec(x+δx) δy=(x+δx)sec(x+δx)−xsecx δy= x+δy cos(x+δx) − x cosx = (x+δx)cosx−xcos(x+δx) cosxcos(x+δx) δy= xcosx+δxcosx−xcos(x+δx) cosxcos(x+δx) δy= xcosx−xcos(x+δx) cosxcos(x+δx) + δxcosx cosxcos(x+δx) δy= x[2sin x+x+δx 2 sin x+δx−x 2 ] cosxcos(x+δx) + δx cos(x+δx) δy= x[2sin(x+ δx 2 )sin δx 2 ] cosxcos(x+δx) + δx cos(x+δx) | Recall cosA−cosB=2sin A+B 2 sin B−A 2 δy= 2xsin(x+ δx 2 )sin δx 2 cosxcos(x+δx) + δx cos(x+δx) Dividing through δy δx = 2xsin(x+ δx 2 )sin δx 2 δxcosxcos(x+δx) + δx δxcos(x+δx) δx δy = xsin(x+ δx 2 ) sin δx 2 δx 2 cosxcos(x+δx) + 1 cos(x+δx) Proceeding to the limit as δx→0 dy dx = lim δx→0 δy δx =x lim δx→0 sin(x+ δx 2 )⋅ lim δx→0 sin δx 2 δx 2 lim δx→0 1 cosxcos(x+δx) + lim δx→0 1 cos(x+δx) dy dx = xsinx cosxcosx + 1 cosx =xtanxsecx+secx d dx (xsecx)=xtanxsecx+secx MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakqaabeqaaiaadMhacq GH9aqpcaWG4bGaci4CaiaacwgacaGGJbGaamiEaaqaaiaadMhacqGH RaWkcqaH0oazcaWG5bGaeyypa0JaaiikaiaadIhacqGHRaWkcqaH0o azcaWG4bGaaiykaiGacohacaGGLbGaai4yaiaacIcacaWG4bGaey4k aSIaeqiTdqMaamiEaiaacMcaaeaacqaH0oazcaWG5bGaeyypa0Jaai ikaiaadIhacqGHRaWkcqaH0oazcaWG4bGaaiykaiGacohacaGGLbGa 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0oazcaWG4bGaaiykaaaaaeaadaWcaaqaaiaadsgacaWG5baabaGaam izaiaadIhaaaGaeyypa0ZaaSaaaeaacaWG4bGaci4CaiaacMgacaGG UbGaamiEaaqaaiGacogacaGGVbGaai4CaiaadIhaciGGJbGaai4Bai aacohacaWG4baaaiabgUcaRmaalaaabaGaaGymaaqaaiGacogacaGG VbGaai4CaiaadIhaaaGaeyypa0JaamiEaiGacshacaGGHbGaaiOBai aadIhaciGGZbGaaiyzaiaacogacaWG4bGaey4kaSIaci4Caiaacwga caGGJbGaamiEaaqaamaalaaabaGaamizaaqaaiaadsgacaWG4baaai aacIcacaWG4bGaci4CaiaacwgacaGGJbGaamiEaiaacMcacqGH9aqp caWG4bGaciiDaiaacggacaGGUbGaamiEaiGacohacaGGLbGaai4yai aadIhacqGHRaWkciGGZbGaaiyzaiaacogacaWG4baaaaa@DE4A@              Determine from first principle the derivative of x e x MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWG4bGaamyzam aaCaaaleqabaGaamiEaaaaaaa@3895@ with respect to x y=x e x y+δy=(x+δx) e (x+δx) δy=(x+δx) e (x+δx) −x e x δy=x e (x+δx) +δx[ e (x+δx) ]−x e x δy=x e (x+δx) −x e x +δx( e (x+δx) ) δy=x[ e (x+δx) − e x ]+δx( e (x+δx) ) δy=x[ e x ⋅ e δx − e x ]+δx( e (x+δx) ) δy=x e x [ e δx −1]+δx( e (x+δx) ) Divide through by δx δy δx = x e x δx ( e δx −1)+ δx( e x+δx ) δx δy δx =x e x ( e δx −1) δx + e x+δx Note:  lim x→0 ( e x −1) x =1 which implies  lim δx→0 ( e δx −1) δx =1 Taking limits as δx→0 dy dx = lim δx→0 δy δx = lim δx→0 x e x ( e δx −1) δx + lim δx→0 e x+δx dy dx =x e x + e x Hence  d dx (x e x )=x e x + e x MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakqaabeqaaiaadMhacq GH9aqpcaWG4bGaamyzamaaCaaaleqabaGaamiEaaaaaOqaaiaadMha cqGHRaWkcqaH0oazcaWG5bGaeyypa0JaaiikaiaadIhacqGHRaWkcq aH0oazcaWG4bGaaiykaiaadwgadaahaaWcbeqaaiaacIcacaWG4bGa ey4kaSIaeqiTdqMaamiEaiaacMcaaaaakeaacqaH0oazcaWG5bGaey ypa0JaaiikaiaadIhacqGHRaWkcqaH0oazcaWG4bGaaiykaiaadwga daahaaWcbeqaaiaacIcacaWG4bGaey4kaSIaeqiTdqMaamiEaiaacM caaaGccqGHsislcaWG4bGaamyzamaaCaaaleqabaGaamiEaaaaaOqa aiabes7aKjaadMhacqGH9aqpcaWG4bGaamyzamaaCaaaleqabaGaai ikaiaadIhacqGHRaWkcqaH0oazcaWG4bGaaiykaaaakiabgUcaRiab es7aKjaadIhacaGGBbGaamyzamaaCaaaleqabaGaaiikaiaadIhacq GHRaWkcqaH0oazcaWG4bGaaiykaaaakiaac2facqGHsislcaWG4bGa amyzamaaCaaaleqabaGaamiEaaaaaOqaaiabes7aKjaadMhacqGH9a qpcaWG4bGaamyzamaaCaaaleqabaGaaiikaiaadIhacqGHRaWkcqaH 0oazcaWG4bGaaiykaaaakiabgkHiTiaadIhacaWGLbWaaWbaaSqabe aacaWG4baaaOGaey4kaSIaeqiTdqMaamiEaiaacIcacaWGLbWaaWba aSqabeaacaGGOaGaamiEaiabgUcaRiabes7aKjaadIhacaGGPaaaaO Gaaiykaaqaaiabes7aKjaadMhacqGH9aqpcaWG4bGaai4waiaadwga daahaaWcbeqaaiaacIcacaWG4bGaey4kaSIaeqiTdqMaamiEaiaacM caaaGccqGHsislcaWGLbWaaWbaaSqabeaacaWG4baaaOGaaiyxaiab gUcaRiabes7aKjaadIhacaGGOaGaamyzamaaCaaaleqabaGaaiikai aadIhacqGHRaWkcqaH0oazcaWG4bGaaiykaaaakiaacMcaaeaacqaH 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aaqabaGcdaWcaaqaaiabes7aKjaadMhaaeaacqaH0oazcaWG4baaai abg2da9maaxababaGaciiBaiaacMgacaGGTbaaleaacqaH0oazcaWG 4bGaeyOKH4QaaGimaaqabaGccaWG4bGaamyzamaaCaaaleqabaGaam iEaaaakmaalaaabaGaaiikaiaadwgadaahaaWcbeqaaiabes7aKjaa dIhaaaGccqGHsislcaaIXaGaaiykaaqaaiabes7aKjaadIhaaaGaey 4kaSYaaCbeaeaaciGGSbGaaiyAaiaac2gaaSqaaiabes7aKjaadIha cqGHsgIRcaaIWaaabeaakiaadwgadaahaaWcbeqaaiaadIhacqGHRa WkcqaH0oazcaWG4baaaaGcbaWaaSaaaeaacaWGKbGaamyEaaqaaiaa dsgacaWG4baaaiabg2da9iaadIhacaWGLbWaaWbaaSqabeaacaWG4b aaaOGaey4kaSIaamyzamaaCaaaleqabaGaamiEaaaaaOqaaiaabIea caqGLbGaaeOBaiaabogacaqGLbGaaeiiamaalaaabaGaamizaaqaai aadsgacaWG4baaaiaacIcacaWG4bGaamyzamaaCaaaleqabaGaamiE aaaakiaacMcacqGH9aqpcaWG4bGaamyzamaaCaaaleqabaGaamiEaa aakiabgUcaRiaadwgadaahaaWcbeqaaiaadIhaaaaaaaa@D6CA@ If y=u+v MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWG5bGaeyypa0 JaamyDaiabgUcaRiaadAhaaaa@3A5F@ where u, v are function of x show that dy dx = du dx + dv dx MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaadaWcaaqaaiaads gacaWG5baabaGaamizaiaadIhaaaGaeyypa0ZaaSaaaeaacaWGKbGa amyDaaqaaiaadsgacaWG4baaaiabgUcaRmaalaaabaGaamizaiaadA haaeaacaWGKbGaamiEaaaaaaa@42FC@ y=u+v y+δy=(u+δu)+(v+δv) δy=(u+δu)+(v+δv)−(u+v) δy=δu+δv Dividing through by δx δy δx = δu δx + δu δx Taking limits as δx→0 dy dx = lim δx→0 ( δu δx + δu δx ) dy dx = du dx + dv dx MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakqaabeqaaiaadMhacq GH9aqpcaWG1bGaey4kaSIaamODaaqaaiaadMhacqGHRaWkcqaH0oaz caWG5bGaeyypa0JaaiikaiaadwhacqGHRaWkcqaH0oazcaWG1bGaai ykaiabgUcaRiaacIcacaWG2bGaey4kaSIaeqiTdqMaamODaiaacMca aeaacqaH0oazcaWG5bGaeyypa0JaaiikaiaadwhacqGHRaWkcqaH0o azcaWG1bGaaiykaiabgUcaRiaacIcacaWG2bGaey4kaSIaeqiTdqMa amODaiaacMcacqGHsislcaGGOaGaamyDaiabgUcaRiaadAhacaGGPa aabaGaeqiTdqMaamyEaiabg2da9iabes7aKjaadwhacqGHRaWkcqaH 0oazcaWG2baabaGaaeiraiaabMgacaqG2bGaaeyAaiaabsgacaqGPb GaaeOBaiaabEgacaqGGaGaaeiDaiaabIgacaqGYbGaae4Baiaabwha caqGNbGaaeiAaiaabccacaqGIbGaaeyEaiaabccacqaH0oazcaWG4b aabaWaaSaaaeaacqaH0oazcaWG5baabaGaeqiTdqMaamiEaaaacqGH 9aqpdaWcaaqaaiabes7aKjaadwhaaeaacqaH0oazcaWG4baaaiabgU caRmaalaaabaGaeqiTdqMaamyDaaqaaiabes7aKjaadIhaaaaabaGa aeivaiaabggacaqGRbGaaeyAaiaab6gacaqGNbGaaeiiaiaabYgaca qGPbGaaeyBaiaabMgacaqG0bGaae4CaiaabccacaqGHbGaae4Caiaa bccacqaH0oazcaWG4bGaeyOKH4QaaGimaaqaamaalaaabaGaamizai aadMhaaeaacaWGKbGaamiEaaaacqGH9aqpdaWfqaqaaiGacYgacaGG PbGaaiyBaaWcbaGaeqiTdqMaamiEaiabgkziUkaaicdaaeqaaOWaae WaaeaadaWcaaqaaiabes7aKjaadwhaaeaacqaH0oazcaWG4baaaiab gUcaRmaalaaabaGaeqiTdqMaamyDaaqaaiabes7aKjaadIhaaaaaca GLOaGaayzkaaaabaWaaSaaaeaacaWGKbGaamyEaaqaaiaadsgacaWG 4baaaiabg2da9maalaaabaGaamizaiaadwhaaeaacaWGKbGaamiEaa aacqGHRaWkdaWcaaqaaiaadsgacaWG2baabaGaamizaiaadIhaaaaa aaa@CDA9@  If y=uvw MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWG5bGaeyypa0 JaamyDaiaadAhacaWG3baaaa@3A79@ show that ^{dy dx =u d dx (vw)+v d dx (uw)+w d dx (uv) MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaadaWcaaqaaiaads gacaWG5baabaGaamizaiaadIhaaaGaeyypa0JaamyDamaalaaabaGa amizaaqaaiaadsgacaWG4baaaiaacIcacaWG2bGaam4DaiaacMcacq GHRaWkcaWG2bWaaSaaaeaacaWGKbaabaGaamizaiaadIhaaaGaaiik aiaadwhacaWG3bGaaiykaiabgUcaRiaadEhadaWcaaqaaiaadsgaae aacaWGKbGaamiEaaaacaGGOaGaamyDaiaadAhacaGGPaaaaa@51A6@}