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zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3inverse cosine, with a little spelling error probably :)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Is this what the problem looks like badria? \[\large y=\arccos\left(\frac{b+a \cos x}{a+b \cos x}\right)\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3So 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
 one year ago
Best ResponseYou've already chosen the best response.3Our 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
 one year ago
Best ResponseYou've already chosen the best response.3\[\large \frac{d}{dx}\arccos \left(\color{royalblue}{\frac{b+a \cos x}{a+b \cos x}}\right)=\frac{1}{\sqrt{1\left(\color{royalblue}{\color{royalblue}{\dfrac{b+a \cos x}{a+b \cos x}}}\right)^2}} \qquad \left(\color{royalblue}{\color{royalblue}{\frac{b+a \cos x}{a+b \cos x}}}\right)'\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Hmm 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
 one year ago
Best ResponseYou've already chosen the best response.3The 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?

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

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

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

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

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

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3\[\large \frac{\color{royalblue}{(b+a \cos x)'}(a+b \cos x)(b+a \cos x)\color{royalblue}{(a+b \cos x)'}}{(a+b \cos x)^2}\] \[\large \frac{\color{orangered}{(0a \sin x)}(a+b \cos x)(b+a \cos x)\color{orangered}{(0+b \sin x)}}{(a+b \cos x)^2}\] Remember, your a's and b's are just constants! They should give you zero when you differentiate them.

AravindG
 one year ago
Best ResponseYou've already chosen the best response.0Hi, 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

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Woops, small error. \[\large \frac{\color{orangered}{(0a \sin x)}(a+b \cos x)(b+a \cos x)\color{orangered}{(0\color{green}{}b \sin x)}}{(a+b \cos x)^2}\]

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

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