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badria
Group Title
Find the derivative
\[y=arcoss \frac{ b+acosx }{ a+bcosx }\]
 one year ago
 one year ago
badria Group Title
Find the derivative \[y=arcoss \frac{ b+acosx }{ a+bcosx }\]
 one year ago
 one year ago

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Azteck Group TitleBest ResponseYou've already chosen the best response.0
what's arcoss?
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.2
inverse cosine, with a little spelling error probably :)
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.2
Is this what the problem looks like badria? \[\large y=\arccos\left(\frac{b+a \cos x}{a+b \cos x}\right)\]
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.2
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.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.2
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.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.2
\[\large \frac{d}{dx}\arccos \left(\color{royalblue}{\frac{b+a \cos x}{a+b \cos x}}\right)=\frac{1}{\sqrt{1\left(\color{royalblue}{\color{royalblue}{\dfrac{b+a \cos x}{a+b \cos x}}}\right)^2}} \qquad \left(\color{royalblue}{\color{royalblue}{\frac{b+a \cos x}{a+b \cos x}}}\right)'\]
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.2
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
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.2
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?
 one year ago

badria Group TitleBest ResponseYou've already chosen the best response.0
Yes i do understand but im a bit confused is on the \[(\frac{ b+acosx }{ a+bcosx })\prime \]
 one year ago

badria Group TitleBest ResponseYou've already chosen the best response.0
would i use the quotient rule ?
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.2
Yes good c:
 one year ago

badria Group TitleBest ResponseYou've already chosen the best response.0
awesome , thanks for your help !
 one year ago

badria Group TitleBest ResponseYou've already chosen the best response.0
Would this be correct \[\frac{ (1+sinx)[(b+acosx)(a+bcosx)] }{ (a+bcosx)^{2} }\]
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
Good work @zepdrix and @badria . And \[\huge{\color{red}{\mathbb{WELCOME}} \space \textbf{TO}\space \color{orange}{\frak{OPENSTUDY}}}\]
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.2
\[\large \frac{\color{royalblue}{(b+a \cos x)'}(a+b \cos x)(b+a \cos x)\color{royalblue}{(a+b \cos x)'}}{(a+b \cos x)^2}\] \[\large \frac{\color{orangered}{(0a \sin x)}(a+b \cos x)(b+a \cos x)\color{orangered}{(0+b \sin x)}}{(a+b \cos x)^2}\] Remember, your a's and b's are just constants! They should give you zero when you differentiate them.
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.0
Hi, I can see that you are a new member here at OpenStudy so, I would first like to say Welcome to OpenStudy, I would like to point you to the chat pods these chat pods are where you can make new friends and talk to new people just like you, I would also like to emphasize our "NONCHEATING POLICY" we ask here at OpenStudy that you do NOT post exam/Test questions if you are caught doing this we will notify your current school, I would also like to ask you to check out the OpenStudy "Code of Conduct" http://openstudy.com/codeofconduct Please read this carefully and thoroughly. Welcome to the OpenStudy Community! Ambassador
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.2
Woops, small error. \[\large \frac{\color{orangered}{(0a \sin x)}(a+b \cos x)(b+a \cos x)\color{orangered}{(0\color{green}{}b \sin x)}}{(a+b \cos x)^2}\]
 one year ago

Azteck Group TitleBest ResponseYou've already chosen the best response.0
What did you get as your answer @badria ?
 one year ago

Azteck Group TitleBest ResponseYou've already chosen the best response.0
This should be your answer I think if you haven't gotten it. \[\frac{d}{dx}\cos^{1}(\frac{b+a\cos x}{a+b\cos x)}=\frac{\sqrt{a^2b^2}}{a+b\cos x}\]
 one year ago
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