# Thread: help in differentiatinr :(

1. ## help in differentiatinr :(

F(x) = {(1+Cosx)^0.5} / {(1-Cosx)^0.5}

2. Originally Posted by Blush
F(x) = {(1+Cosx)^0.5} / {(1-Cosx)^0.5}

are u asking for F' ?

3. yeah.. and thanx alot for being concern n for all ur answers

4. ${d\over dx}\biggl((1+\cos x)^{.5} (1-\cos x)^{-.5}\biggr)$

$=.5(1+\cos x)^{-.5}(-\sin x) (1-\cos x)^{-.5}+(1+\cos x)^{.5}(-.5)(1-\cos x)^{-1.5}(\sin x)$

5. Keeping the function as a quotient and using quotient rule also works

$u(x) = (1+\cos x)^{\frac{1}{2}}$
$v(x) = (1-\cos x)^{\frac{1}{2}}$

$f'(x) = \frac{vu' - uv'}{v^2}$

Just remember the chain rule when differentiating u and v

6. Originally Posted by Blush
F(x) = {(1+Cosx)^0.5} / {(1-Cosx)^0.5}

y=(-(1 - cos x)^(1/2)*1/2*(1 + cos x)^(-1/2)*sin x - (1 + cos x)^(1/2)*1/2*(1 - cos x)^(-1/2)*sin x)/((1 - cos x)^(1/2)^2)

I calculated it using a program. I would use the chain rule and quotient rule. But, I'm not sure if there's an identity you can use...

By the way, how do you guys write the equations using the fancy d/dx and stuff...

7. Originally Posted by econstudent

y=(-(1 - cos x)^(1/2)*1/2*(1 + cos x)^(-1/2)*sin x - (1 + cos x)^(1/2)*1/2*(1 - cos x)^(-1/2)*sin x)/((1 - cos x)^(1/2)^2)

I calculated it using a program. I would use the chain rule and quotient rule. But, I'm not sure if there's an identity you can use...

By the way, how do you guys write the equations using the fancy d/dx and stuff...
$\frac{d}{dx}\left(f(x)\right)$ just means the differential of $f(x)$ with respect to x

8. Originally Posted by econstudent
[snip]

By the way, how do you guys write the equations using the fancy d/dx and stuff...
Read what's in this subforum: http://www.mathhelpforum.com/math-help/latex-help/

9. Originally Posted by mr fantastic
Haha, I completely misinterpreted what he was asking!