Derivative of inverse tangent function

Homework Statement

Find derivative of tan$^{-1}$($\frac{3sinx}{4+5cosx}$)

Homework Equations

deriviative of tan$^{-1}$=$\frac{U'}{1+U^{2}}$

The Attempt at a Solution

I found U'= $\frac{12cosx+15}{(4+5cosx)^{2}}$

1+U$^{2}$=1+$\frac{9sin^{2}x}{(4+5cosx)^{2}}$

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)?

The Attempt at a Solution

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tiny-tim
Homework Helper
hi biochem850!

looks ok so far

How do you then simplify:

$\frac{U'}{1+U^{2}}$?

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

tiny-tim
Homework Helper
hi biochem850!

(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!

$\frac{12cosx+15}{(4+5cosx)^{2}}$*1+$\frac{(4+5cosx)^{2}}{9sin^{2}x}$=

$\frac{12cosx+15(9sin^{2}x)+(4+5cosx)^{2}(4+5cosx)^{2}}{(4+5cosx)^{2}(9sin^{2}x)}$

I'm not sure this is correct.

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tiny-tim
Homework Helper
hmm … let's use tex instead of itex, to make it bigger
$$\frac{12cosx+15}{(4+5cosx)^{2}} \left(1+\frac{(4+5cosx)^{2}}{9sin^{2}x}\right)$$=

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

the bracket is 1 + 1/U2, it should be 1/(1 + U2)

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
Homework Helper
I changed my work just before you posted
but you didn't change the bit i said was wrong

but you didn't change the bit i said was wrong
$\frac{12cosx+15}{(4+5cosx)^{2}}$$\frac{(4+5cosx)^{2}+1}{9sin^{2}x+1}$=

$\frac{12cosx+15}{9sin^{2}x+1}$
I'm not sure this is correct. I really don't understand what you asked me to change.

tiny-tim
Homework Helper
change your original …
1+$\frac{9sin^{2}x}{(4+5cosx)^{2}}$
[/SIZE]
… to something with (4 + 5cosx)2 on the bottom

change your original …

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

$\frac{(4+5cosx)^{2}+9sin^{2}x}{(4+5cosx)^{2}}$?

tiny-tim
Homework Helper
$\frac{(4+5cosx)^{2}+9sin^{2}x}{(4+5cosx)^{2}}$?
yup!

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

$\frac{12cosx+15}{(4+5cosx)^{2}+9sin^{2}x}$

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
Homework Helper
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! :zzz:​

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! :zzz:​
Through some weird algebraic manipulation I simplified down to $\frac{3}{5+4cosx}$. I beleive this is correct. I'm going to sleep.

Thanks so much!