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badria

  • 2 years ago

Find the derivative \[y=arcoss \frac{ b+acosx }{ a+bcosx }\]

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  1. Azteck
    • 2 years ago
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    what's arcoss?

  2. zepdrix
    • 2 years ago
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    inverse cosine, with a little spelling error probably :)

  3. zepdrix
    • 2 years ago
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    Is this what the problem looks like badria? \[\large y=\arccos\left(\frac{b+a \cos x}{a+b \cos x}\right)\]

  4. badria
    • 2 years ago
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    Yes

  5. zepdrix
    • 2 years ago
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    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.

  6. zepdrix
    • 2 years ago
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    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.

  7. zepdrix
    • 2 years ago
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    \[\large \frac{d}{dx}\arccos \left(\color{royalblue}{\frac{b+a \cos x}{a+b \cos x}}\right)=\frac{-1}{\sqrt{1-\left(\color{royalblue}{\color{royalblue}{\dfrac{b+a \cos x}{a+b \cos x}}}\right)^2}} \qquad \left(\color{royalblue}{\color{royalblue}{\frac{b+a \cos x}{a+b \cos x}}}\right)'\]

  8. zepdrix
    • 2 years ago
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    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

  9. zepdrix
    • 2 years ago
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    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?

  10. badria
    • 2 years ago
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    Yes i do understand but im a bit confused is on the \[(\frac{ b+acosx }{ a+bcosx })\prime \]

  11. badria
    • 2 years ago
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    would i use the quotient rule ?

  12. zepdrix
    • 2 years ago
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    Yes good c:

  13. badria
    • 2 years ago
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    awesome , thanks for your help !

  14. badria
    • 2 years ago
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    Would this be correct \[\frac{ (1+sinx)[(b+acosx)-(a+bcosx)] }{ (a+bcosx)^{2} }\]

  15. mathslover
    • 2 years ago
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    Good work @zepdrix and @badria . And \[\huge{\color{red}{\mathbb{WELCOME}} \space \textbf{TO}\space \color{orange}{\frak{OPENSTUDY}}}\]

  16. badria
    • 2 years ago
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    thank you

  17. badria
    • 2 years ago
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    @zepdrix

  18. zepdrix
    • 2 years ago
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    \[\large \frac{\color{royalblue}{(b+a \cos x)'}(a+b \cos x)-(b+a \cos x)\color{royalblue}{(a+b \cos x)'}}{(a+b \cos x)^2}\] \[\large \frac{\color{orangered}{(0-a \sin x)}(a+b \cos x)-(b+a \cos x)\color{orangered}{(0+b \sin x)}}{(a+b \cos x)^2}\] Remember, your a's and b's are just constants! They should give you zero when you differentiate them.

  19. AravindG
    • 2 years ago
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    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 "NON-CHEATING 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/code-of-conduct Please read this carefully and thoroughly. Welcome to the OpenStudy Community! Ambassador

  20. zepdrix
    • 2 years ago
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    Woops, small error. \[\large \frac{\color{orangered}{(0-a \sin x)}(a+b \cos x)-(b+a \cos x)\color{orangered}{(0\color{green}{-}b \sin x)}}{(a+b \cos x)^2}\]

  21. badria
    • 2 years ago
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    Thanks !

  22. Azteck
    • 2 years ago
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    What did you get as your answer @badria ?

  23. Azteck
    • 2 years ago
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    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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