Derivative of inverse tangent function

In summary: This has been really helpful!In summary, the person was trying to find the derivative of tan^{-1}(\frac{3sinx}{4+5cosx}) but they were having trouble and needed help. They found that \frac{3}{5+4cosx} was the derivative and they went to sleep.
  • #1
biochem850
51
0

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:
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  • #2
hi biochem850! :smile:

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

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

I've tried to simplify this but with no luck.
 
  • #4
Can someone please help me?
 
  • #5
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
[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
hmm :rolleyes: … let's use tex instead of itex, to make it bigger :wink:
biochem850 said:
[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
tiny-tim said:
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
biochem850 said:
I changed my work just before you posted

but you didn't change the bit i said was wrong :confused:
 
  • #10
tiny-tim said:
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
change your original …
biochem850 said:
1+[itex]\frac{9sin^{2}x}{(4+5cosx)^{2}}[/itex]
[/SIZE]

… to something with (4 + 5cosx)2 on the bottom
 
  • #12
tiny-tim said:
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
biochem850 said:
[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
[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
biochem850 said:
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
tiny-tim said:
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 believe this is correct. I'm going to sleep.:smile:

Thanks so much!
 

What is the derivative of inverse tangent function?

The derivative of the inverse tangent function, also known as arctangent or tan-1, is given by the formula 1 / (1 + x2).

What is the domain and range of the inverse tangent function?

The domain of the inverse tangent function is all real numbers, while the range is between -π/2 and π/2, or -90° and 90°.

How do you find the derivative of inverse tangent function?

The derivative of inverse tangent function can be found by using the formula d/dx(tan-1(x)) = 1 / (1 + x2).

What is the relationship between the derivative of inverse tangent function and the derivative of tangent function?

The derivative of inverse tangent function is the inverse of the derivative of tangent function. This means that if the derivative of tangent function is 1 / cos2(x), then the derivative of inverse tangent function is 1 / (1 + x2).

What are some real-world applications of the derivative of inverse tangent function?

The derivative of inverse tangent function is used in physics, engineering, and geometry to calculate the slope of a curve at a specific point, as well as to find the rate of change of a variable over time.

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