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

zepdrixBest 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

zepdrixBest 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

zepdrixBest 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

zepdrixBest 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

zepdrixBest 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

zepdrixBest 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

badriaBest 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

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

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

badriaBest 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

mathsloverBest 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

zepdrixBest 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

AravindGBest ResponseYou've already chosen the best response.0
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 one year ago

zepdrixBest 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

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

AzteckBest 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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