B Buzzins May 2010 8 0 May 12, 2010 #1 okay so my previous question was answered however i know that: 2cos(x).cos(9x) has the derivative: -18cos(x)sin(9x) - 2sin(x)cos(9x) But will 2[cos1/2(f1-f2)t][cos1/2(f1+f2)t] have the same derivative? with f1= 10 and f2=8
okay so my previous question was answered however i know that: 2cos(x).cos(9x) has the derivative: -18cos(x)sin(9x) - 2sin(x)cos(9x) But will 2[cos1/2(f1-f2)t][cos1/2(f1+f2)t] have the same derivative? with f1= 10 and f2=8
B Buzzins May 2010 8 0 May 12, 2010 #3 hrmmm. that makes it all seem a bit too easy then. i am meant to verify the answer from the first equation with that one. But to me they are basically the same.
hrmmm. that makes it all seem a bit too easy then. i am meant to verify the answer from the first equation with that one. But to me they are basically the same.
pickslides MHF Helper Sep 2008 5,237 1,625 Melbourne May 12, 2010 #4 They are the same. Just find the derivavtive of the second to confirm.
B Buzzins May 2010 8 0 May 12, 2010 #5 so from; y=2[cos1/2(10-8)x][cos1/2(10+8)x] = 2[cos1/2(2)x][cos1/2(18)x] =2cosx.cos(9x) then differentiate that and it should equal -18cos(x)sin(9x)-2sin(x)cos(9x) thats like two extra steps rofl.
so from; y=2[cos1/2(10-8)x][cos1/2(10+8)x] = 2[cos1/2(2)x][cos1/2(18)x] =2cosx.cos(9x) then differentiate that and it should equal -18cos(x)sin(9x)-2sin(x)cos(9x) thats like two extra steps rofl.
pickslides MHF Helper Sep 2008 5,237 1,625 Melbourne May 12, 2010 #6 Exactly (Rofl) . You were using a previous result to help you with another. This will happen a lot in your studies. Reactions: Buzzins
Exactly (Rofl) . You were using a previous result to help you with another. This will happen a lot in your studies.