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Re: Nairaland Mathematics Clinic by killsmith(m): 8:41pm On May 26 
tannyp:Hope the exam was a success? 
Re: Nairaland Mathematics Clinic by GMacbeth: 6:15pm On Jun 02 
Please help me out on this

Re: Nairaland Mathematics Clinic by MeetDx(m): 11:05am On Jun 04 
^^^^ Solution to Q1 If y = tan(2x) Using chain rule => Let's Put u = 2x, then... du/dx = 2 so that y = tanu and dy/du = sec^{2}u :. dy/dx = dy/du* du/dx => sec^{2}(2x) * 2 :. The derivative of tan(2x) => y' => d/dx{tan2x} => 2sec^{2}(2x) =>2/cos^{2}(2x) To show that y' = 2(1 + y^{2}) Take from the RHS 2(1 + y^{2}) => 2{1 + tan^{2}(2x)} => 2{1 + sin^{2}(2x)]/cos^{2}(2x)} => 2{cos^{2}(2x) + Sin^{2}(2x)}/ cos^{2}(2x). Since sin^{2}A + Cos^{2}A = 1, then => cos^{2}(2x) + Sin^{2}(2x) => 1 :. 2(1 + y^{2}) => 2{1/cos^{2}(2x)} => 2sec^{2}(2x) => y' => LHS 
Re: Nairaland Mathematics Clinic by MeetDx(m): 11:32am On Jun 04 
GMacbeth: Solution to Q2 y = cot(2x) Using chain rule let's put u = 2x So that du/dx = 2 and y = cotu :. dy/du = cosec^{2}u dy/dx = dy/du * du/dx => cosec^{2}u * 2 => 2cosec^{2}u y'=> 2cosec^{2}(2x) To show that y' = 2( 1 + y^{2}) From RHS => substitute y = cot(2x) =>  2{ 1 + cot^{2}(2x)} => 2{1 + cos^{2}(2x)/sin^{2}(2x)} => 2{ sin^{2}(2x) + cos^{2}(2x)}/sin^{2}(2x) Since Sin^{2}A + Cos^{2}A = 1 sin^{2}(2x) + cos^{2}(2x) = 1 :. 2{ 1/sin^{2}(2x)} => 2cosec^{2}(2x) => y' => LHS 
Re: Nairaland Mathematics Clinic by Blessingposhh: 9:22am On Jun 10 
Please sir help me out on this

Re: Nairaland Mathematics Clinic by Ragnarr(m): 7:12am On Jun 11 
Please help on this

Re: Nairaland Mathematics Clinic by mathefaro(m): 7:36am On Jun 11 
Ragnarr:we know that Sin²x + Cos²x = 1 Hence cos²x = 1  Sin²x Therefore, the LHS of the equation can become sqrt(Cos²x) = Cos x = RHS 
Re: Nairaland Mathematics Clinic by Ragnarr(m): 1:30pm On Jun 11 
mathefaro:thank u sir 1 Like 
Re: Nairaland Mathematics Clinic by cole01: 6:54am On Jun 13 
Mathefano MeetDx mechanics96 Yusman14 Killsmith Lapacian All the bosses in the house, plz I really need to submit this, and I just found this thread out, plz help out

Re: Nairaland Mathematics Clinic by SPLENDID25(m): 6:23pm On Jun 13 
Please help me out with these.

Re: Nairaland Mathematics Clinic by Flyingngel(m): 9:53pm On Jun 14 
Am grateful for stumbling on a thread like this. I knw it is not late for me. 
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