1. The problem statement, all variables and given/known data Find derivative of tan[itex]^{-1}[/itex]([itex]\frac{3sinx}{4+5cosx}[/itex]) 2. Relevant equations deriviative of tan[itex]^{-1}[/itex]=[itex]\frac{U'}{1+U^{2}}[/itex] 3. The attempt at a solution I found U'= [itex]\frac{12cosx+15}{(4+5cosx)^{2}}[/itex] 1+U[itex]^{2}[/itex]=1+[itex]\frac{9sin^{2}x}{(4+5cosx)^{2}}[/itex] I think my components are correct but my answer is still incorrect. Are these basic components even correct (if so I can proceed on my own from here)? 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution
How do you then simplify: [itex]\frac{U'}{1+U^{2}}[/itex]? I've tried to simplify this but with no luck.
hi biochem850! (just got up :zzz:) well, the (4 + 5cosx)^{2} should cancel and disappear … show us your full calculations, and then we'll know how to help!
[itex]\frac{12cosx+15}{(4+5cosx)^{2}}[/itex]*1+[itex]\frac{(4+5cosx)^{2}}{9sin^{2}x}[/itex]= [itex]\frac{12cosx+15(9sin^{2}x)+(4+5cosx)^{2}(4+5cosx)^{2}}{(4+5cosx)^{2}(9sin^{2}x)}[/itex] I'm not sure this is correct.
hmm … let's use tex instead of itex, to make it bigger … no, that first line is wrong, the bracket is 1 + 1/U^{2}, it should be 1/(1 + U^{2})
[itex]\frac{12cosx+15}{(4+5cosx)^{2}}[/itex][itex]\frac{(4+5cosx)^{2}+1}{9sin^{2}x+1}[/itex]= [itex]\frac{12cosx+15}{9sin^{2}x+1}[/itex] I'm not sure this is correct. I really don't understand what you asked me to change.
[itex]\frac{12cosx+15}{(4+5cosx)^{2}+9sin^{2}x}[/itex] In terms of finding the derivative I know I've found it but this can be simplified further. I don't know if I'm tired or what because I cannot seem to do these very simple problems. This does not bode well for my calculus mark. Would you factor out a 3 in the numerator and then see what will simplify?
no, and i'd be very surprised if this simplifies (except that you could get rid of the sin^{2} by using cos^{2} + sin^{2} = 1 ) get some sleep! :zzz:
Through some weird algebraic manipulation I simplified down to [itex]\frac{3}{5+4cosx}[/itex]. I beleive this is correct. I'm going to sleep. Thanks so much!