## badria Find the derivative $y=arcoss \frac{ b+acosx }{ a+bcosx }$ 11 months ago 11 months ago

1. Azteck

what's arcoss?

2. zepdrix

inverse cosine, with a little spelling error probably :)

3. zepdrix

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

Yes

5. zepdrix

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

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

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

8. zepdrix

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

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 ?

12. zepdrix

Yes good c:

awesome , thanks for your help !

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

15. mathslover

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

thank you

@zepdrix

18. zepdrix

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

19. AravindG

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20. zepdrix

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

Thanks !

22. Azteck

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