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Derivative of inverse tangent function

by biochem850
Tags: derivative, function, inverse, tangent
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biochem850
#1
Mar8-12, 04:53 PM
P: 46
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
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tiny-tim
#2
Mar8-12, 05:10 PM
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hi biochem850!

looks ok so far
biochem850
#3
Mar8-12, 05:39 PM
P: 46
How do you then simplify:

[itex]\frac{U'}{1+U^{2}}[/itex]?

I've tried to simplify this but with no luck.

biochem850
#4
Mar9-12, 02:55 AM
P: 46
Derivative of inverse tangent function

Can someone please help me?
tiny-tim
#5
Mar9-12, 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!
biochem850
#6
Mar9-12, 03:40 AM
P: 46
[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.
tiny-tim
#7
Mar9-12, 03:59 AM
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hmm let's use tex instead of itex, to make it bigger
Quote Quote by biochem850 View Post
[tex]\frac{12cosx+15}{(4+5cosx)^{2}} \left(1+\frac{(4+5cosx)^{2}}{9sin^{2}x}\right)[/tex]=

[tex]\frac{12cosx+15}{9sin^{2}x}[/tex]=[tex]\frac{3(4cosx+5)}{9sin^{2}x}[/tex]
no, that first line is wrong,

the bracket is 1 + 1/U2, it should be 1/(1 + U2)
biochem850
#8
Mar9-12, 04:02 AM
P: 46
Quote Quote by tiny-tim View Post
hmm let's use tex instead of itex, to make it bigger

no, that first line is wrong,

the bracket is 1 + 1/U2, it should be 1/(1 + U2)
I changed my work just before you posted
tiny-tim
#9
Mar9-12, 04:11 AM
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Quote Quote by biochem850 View Post
I changed my work just before you posted
but you didn't change the bit i said was wrong
biochem850
#10
Mar9-12, 04:22 AM
P: 46
Quote Quote by tiny-tim View Post
but you didn't change the bit i said was wrong
[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.
tiny-tim
#11
Mar9-12, 04:33 AM
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change your original
Quote Quote by biochem850 View Post
1+[itex]\frac{9sin^{2}x}{(4+5cosx)^{2}}[/itex]
[/SIZE]
to something with (4 + 5cosx)2 on the bottom
biochem850
#12
Mar9-12, 04:47 AM
P: 46
Quote Quote by tiny-tim View Post
change your original

to something with (4 + 5cosx)2 on the bottom


[itex]\frac{(4+5cosx)^{2}+9sin^{2}x}{(4+5cosx)^{2}}[/itex]?
tiny-tim
#13
Mar9-12, 04:54 AM
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Quote Quote by biochem850 View Post
[itex]\frac{(4+5cosx)^{2}+9sin^{2}x}{(4+5cosx)^{2}}[/itex]?
yup!

now turn that upside-down, and multiply, and the (4 + 5cosx)2 should cancel
biochem850
#14
Mar9-12, 05:26 AM
P: 46
[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?
tiny-tim
#15
Mar9-12, 05:30 AM
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Quote Quote by biochem850 View Post
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 sin2 by using cos2 + sin2 = 1 )
get some sleep!
biochem850
#16
Mar9-12, 05:47 AM
P: 46
Quote Quote by tiny-tim View Post
no, and i'd be very surprised if this simplifies

(except that you could get rid of the sin2 by using cos2 + sin2 = 1 )
get some sleep!
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!


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