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Derivative of inverse tangent function

  • Thread starter biochem850
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  • #1
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Homework Statement



Find derivative of tan[itex]^{-1}[/itex]([itex]\frac{3sinx}{4+5cosx}[/itex])

Homework Equations



deriviative of tan[itex]^{-1}[/itex]=[itex]\frac{U'}{1+U^{2}}[/itex]



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)?

Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution

 
Last edited:

Answers and Replies

  • #2
tiny-tim
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hi biochem850! :smile:

looks ok so far
 
  • #3
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How do you then simplify:

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

I've tried to simplify this but with no luck.
 
  • #4
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Can someone please help me?
 
  • #5
tiny-tim
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hi biochem850! :wink:

(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! :smile:
 
  • #6
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[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.
 
Last edited:
  • #7
tiny-tim
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hmm :rolleyes: … let's use tex instead of itex, to make it bigger :wink:
[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)
 
  • #8
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hmm :rolleyes: … let's use tex instead of itex, to make it bigger :wink:

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
 
  • #9
tiny-tim
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I changed my work just before you posted
but you didn't change the bit i said was wrong :confused:
 
  • #10
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but you didn't change the bit i said was wrong :confused:
[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.
 
  • #11
tiny-tim
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change your original …
1+[itex]\frac{9sin^{2}x}{(4+5cosx)^{2}}[/itex]
[/SIZE]
… to something with (4 + 5cosx)2 on the bottom
 
  • #12
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change your original …

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


[itex]\frac{(4+5cosx)^{2}+9sin^{2}x}{(4+5cosx)^{2}}[/itex]?
 
  • #13
tiny-tim
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[itex]\frac{(4+5cosx)^{2}+9sin^{2}x}{(4+5cosx)^{2}}[/itex]?
yup! :smile:

now turn that upside-down, and multiply, and the (4 + 5cosx)2 should cancel :wink:
 
  • #14
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[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
tiny-tim
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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 :wink:)

get some sleep! :zzz:​
 
  • #16
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no, and i'd be very surprised if this simplifies

(except that you could get rid of the sin2 by using cos2 + sin2 = 1 :wink:)

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.:smile:

Thanks so much!
 

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