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Derivative of inverse tangent function 
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#1
Mar812, 04:53 PM

P: 48

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 


#3
Mar812, 05:39 PM

P: 48

How do you then simplify:
[itex]\frac{U'}{1+U^{2}}[/itex]? I've tried to simplify this but with no luck. 


#4
Mar912, 02:55 AM

P: 48

Derivative of inverse tangent function
Can someone please help me?



#5
Mar912, 03:13 AM

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hi biochem850!
(just got up ) well, the (4 + 5cosx)^{2} should cancel and disappear … show us your full calculations, and then we'll know how to help! 


#6
Mar912, 03:40 AM

P: 48

[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}(9s in^{2}x)}[/itex] I'm not sure this is correct. 


#7
Mar912, 03:59 AM

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hmm … let's use tex instead of itex, to make it bigger …
the bracket is 1 + 1/U^{2}, it should be 1/(1 + U^{2}) 


#8
Mar912, 04:02 AM

P: 48




#9
Mar912, 04:11 AM

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P: 26,148




#10
Mar912, 04:22 AM

P: 48

[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. 


#11
Mar912, 04:33 AM

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change your original …



#12
Mar912, 04:47 AM

P: 48

[itex]\frac{(4+5cosx)^{2}+9sin^{2}x}{(4+5cosx)^{2}}[/itex]? 


#13
Mar912, 04:54 AM

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now turn that upsidedown, and multiply, and the (4 + 5cosx)^{2} should cancel 


#14
Mar912, 05:26 AM

P: 48

[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? 


#15
Mar912, 05:30 AM

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(except that you could get rid of the sin^{2} by using cos^{2} + sin^{2} = 1 ) get some sleep! 


#16
Mar912, 05:47 AM

P: 48

Thanks so much! 


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