Question

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

Answer 1

what's arcoss?

Answer 2

inverse cosine, with a little spelling error probably :)

Answer 3

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

Answer 4

Yes

Answer 5

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.

Answer 6

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.

Answer 7

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

Answer 8

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

Answer 9

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?

Answer 10

Yes i do understand but im a bit confused is on the \[(\frac{ b+acosx }{ a+bcosx })\prime \]

Answer 11

would i use the quotient rule ?

Answer 12

Yes good c:

Answer 13

awesome , thanks for your help !

Answer 14

Would this be correct
\[\frac{ (1+sinx)[(b+acosx)-(a+bcosx)] }{ (a+bcosx)^{2} }\]

Answer 15

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

Answer 16

thank you

Answer 17

@zepdrix

Answer 18

\[\large \frac{\color{royalblue}{(b+a \cos x)'}(a+b \cos x)-(b+a \cos x)\color{royalblue}{(a+b \cos x)'}}{(a+b \cos x)^2}\]

\[\large \frac{\color{orangered}{(0-a \sin x)}(a+b \cos x)-(b+a \cos x)\color{orangered}{(0+b \sin x)}}{(a+b \cos x)^2}\]

Remember, your a's and b's are just constants! They should give you zero when you differentiate them.

Answer 19

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Answer 20

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

Answer 21

Thanks !

Answer 22

What did you get as your answer @badria ?

Answer 23

This should be your answer I think if you haven't gotten it.
\[\frac{d}{dx}\cos^{-1}(\frac{b+a\cos x}{a+b\cos x)}=-\frac{\sqrt{a^2-b^2}}{a+b\cos x}\]