Finding the derivative of y

Question

If $$y= \arccos \frac{2x-3\sqrt{1-x^2}}{\sqrt{13}}$$ ,find $\frac{dy}{dx}.$

Please help me solve this.
Thank you

I think you should learn from your previous question like this. Please see @Brian's answer again. — Mikasa, Aug 25 '13 at 06:57
I agree with @Babak: this is very much like your previous question. — Brian M. Scott, Aug 25 '13 at 06:58
Id like to get to the answer Without Directly applying chain rule — chndn, Aug 25 '13 at 07:03

score 3 · Answer 1 · answered Aug 25 '13 at 06:58

3

HINT:

Putting $x=\cos B, 2=r\cos A,3=r\sin A$ where $r>0$

$(r\cos A)^2+(r\sin A)^2=2^2+3^2=13$

$\implies r=\sqrt{13}$ as $r>0$

and $\cos A=\frac2{\sqrt{13}},\sin A=\frac3{\sqrt{13}}$

$$\implies \frac{2x-3\sqrt{1-x^2}}{\sqrt{13}}=\cos B\cos A-\sin A\sin B=\cos(A+B)=\cos \left(\arccos\frac2{\sqrt{13}}+\arccos x\right)$$

$$\implies \arccos\left(\frac{2x-3\sqrt{1-x^2}}{\sqrt{13}}\right)=\arccos\frac2{\sqrt{13}}+\arccos x$$

answered Aug 25 '13 at 06:58

lab bhattacharjee

+1 However, personally, I preferred you leave some parts for the OP unsolved or hidden. :-) – Mikasa Aug 25 '13 at 07:03
@BabakS., the last line ? :) – lab bhattacharjee Aug 25 '13 at 07:04
It can be, lab. :-) – Mikasa Aug 25 '13 at 07:05

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