anonymous
  • anonymous
Find the derivative \[y=arcoss \frac{ b+acosx }{ a+bcosx }\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
what's arcoss?
zepdrix
  • zepdrix
inverse cosine, with a little spelling error probably :)
zepdrix
  • zepdrix
Is this what the problem looks like badria? \[\large y=\arccos\left(\frac{b+a \cos x}{a+b \cos x}\right)\]

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anonymous
  • anonymous
Yes
zepdrix
  • zepdrix
So let's recall the derivative of inverse cosine. \[\large \frac{d}{dx}\arccos \color{royalblue}{x}=\frac{-1}{\sqrt{1-\color{royalblue}{x}^2}}\] Hopefully that looks correct.
zepdrix
  • zepdrix
Our problem will work out very similarly. Since the inside of our function is more than just \(\large x\), we have to apply the chain rule. It will get a little messy.
zepdrix
  • zepdrix
\[\large \frac{d}{dx}\arccos \left(\color{royalblue}{\frac{b+a \cos x}{a+b \cos x}}\right)=\frac{-1}{\sqrt{1-\left(\color{royalblue}{\color{royalblue}{\dfrac{b+a \cos x}{a+b \cos x}}}\right)^2}} \qquad \left(\color{royalblue}{\color{royalblue}{\frac{b+a \cos x}{a+b \cos x}}}\right)'\]
zepdrix
  • zepdrix
Hmm is there perhaps some identity we're suppose to use before differentiating this? I mean this will lead us in the right direction, it just won't be pretty... hmm
zepdrix
  • zepdrix
The term with the prime on it showed up due to the chain rule. We still need to differentiate that part. Understand what I did there? Or need a little explanation?
anonymous
  • anonymous
Yes i do understand but im a bit confused is on the \[(\frac{ b+acosx }{ a+bcosx })\prime \]
anonymous
  • anonymous
would i use the quotient rule ?
zepdrix
  • zepdrix
Yes good c:
anonymous
  • anonymous
awesome , thanks for your help !
anonymous
  • anonymous
Would this be correct \[\frac{ (1+sinx)[(b+acosx)-(a+bcosx)] }{ (a+bcosx)^{2} }\]
mathslover
  • mathslover
Good work @zepdrix and @badria . And \[\huge{\color{red}{\mathbb{WELCOME}} \space \textbf{TO}\space \color{orange}{\frak{OPENSTUDY}}}\]
anonymous
  • anonymous
thank you
anonymous
  • anonymous
@zepdrix
zepdrix
  • zepdrix
\[\large \frac{\color{royalblue}{(b+a \cos x)'}(a+b \cos x)-(b+a \cos x)\color{royalblue}{(a+b \cos x)'}}{(a+b \cos x)^2}\] \[\large \frac{\color{orangered}{(0-a \sin x)}(a+b \cos x)-(b+a \cos x)\color{orangered}{(0+b \sin x)}}{(a+b \cos x)^2}\] Remember, your a's and b's are just constants! They should give you zero when you differentiate them.
AravindG
  • AravindG
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zepdrix
  • zepdrix
Woops, small error. \[\large \frac{\color{orangered}{(0-a \sin x)}(a+b \cos x)-(b+a \cos x)\color{orangered}{(0\color{green}{-}b \sin x)}}{(a+b \cos x)^2}\]
anonymous
  • anonymous
Thanks !
anonymous
  • anonymous
What did you get as your answer @badria ?
anonymous
  • anonymous
This should be your answer I think if you haven't gotten it. \[\frac{d}{dx}\cos^{-1}(\frac{b+a\cos x}{a+b\cos x)}=-\frac{\sqrt{a^2-b^2}}{a+b\cos x}\]

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