Here's the question you clicked on:
badria
Find the derivative \[y=arcoss \frac{ b+acosx }{ a+bcosx }\]
inverse cosine, with a little spelling error probably :)
Is this what the problem looks like badria? \[\large y=\arccos\left(\frac{b+a \cos x}{a+b \cos x}\right)\]
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.
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.
\[\large \frac{d}{dx}\arccos \left(\color{royalblue}{\frac{b+a \cos x}{a+b \cos x}}\right)=\frac{-1}{\sqrt{1-\left(\color{royalblue}{\color{royalblue}{\dfrac{b+a \cos x}{a+b \cos x}}}\right)^2}} \qquad \left(\color{royalblue}{\color{royalblue}{\frac{b+a \cos x}{a+b \cos x}}}\right)'\]
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
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?
Yes i do understand but im a bit confused is on the \[(\frac{ b+acosx }{ a+bcosx })\prime \]
would i use the quotient rule ?
awesome , thanks for your help !
Would this be correct \[\frac{ (1+sinx)[(b+acosx)-(a+bcosx)] }{ (a+bcosx)^{2} }\]
Good work @zepdrix and @badria . And \[\huge{\color{red}{\mathbb{WELCOME}} \space \textbf{TO}\space \color{orange}{\frak{OPENSTUDY}}}\]
\[\large \frac{\color{royalblue}{(b+a \cos x)'}(a+b \cos x)-(b+a \cos x)\color{royalblue}{(a+b \cos x)'}}{(a+b \cos x)^2}\] \[\large \frac{\color{orangered}{(0-a \sin x)}(a+b \cos x)-(b+a \cos x)\color{orangered}{(0+b \sin x)}}{(a+b \cos x)^2}\] Remember, your a's and b's are just constants! They should give you zero when you differentiate them.
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Woops, small error. \[\large \frac{\color{orangered}{(0-a \sin x)}(a+b \cos x)-(b+a \cos x)\color{orangered}{(0\color{green}{-}b \sin x)}}{(a+b \cos x)^2}\]
What did you get as your answer @badria ?
This should be your answer I think if you haven't gotten it. \[\frac{d}{dx}\cos^{-1}(\frac{b+a\cos x}{a+b\cos x)}=-\frac{\sqrt{a^2-b^2}}{a+b\cos x}\]