Derivative of inverse tangent function

biochem850

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

tiny-tim

hi biochem850!

looks ok so far

biochem850

How do you then simplify:

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

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

biochem850

Can someone please help me?

tiny-tim

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

[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

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)

biochem850

I changed my work just before you posted

tiny-tim

but you didn't change the bit i said was wrong

biochem850

[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

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

biochem850

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

tiny-tim

yup!

now turn that upside-down, and multiply, and the (4 + 5cosx)² should cancel

biochem850

[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

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

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!