Question

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

Answer 1

Sol:

$y={\cos }^{-1}\left\{\frac{2x-3\sqrt{1-x^2}}{\sqrt{13}}\right\}$, To find $\frac{dy}{dx}$

put $x=\sin \theta$, Hence $\theta =\sin +x$

$y={\cos }^{-1}\left\{\frac{2\sin \theta -3\sqrt{1-{\sin }^2\theta }}{\sqrt{13}}\right\}$

$(1-{\sin }^2\theta ={\cos }^2\theta )(since{\sin }^2\theta +\cos \theta =1)$

$y={\cos }^{-1}\left\{\frac{2\sin 6-3\sqrt{{\cos }^{2}6}}{\sqrt{13}}\right\}$

$y={\cos }^{-1}\left\{\frac{2\sin 6-3\cos 6}{\sqrt{13}}\right\}({\cos }^{2}6=\cos 6)$

$y={\cos }^{-1}\{\frac{2}{\sqrt{13}}\sin 6-\frac{2}{\sqrt{13}}\cos \theta \}$

Now get $\frac{2}{\sqrt{13}}=\cos \alpha ,2\frac{3}{\sqrt{13}}=\sin \alpha$

$[{\sin }^2\alpha +{\cos }^2\alpha ={\left(\frac{2}{\sqrt{13}}\right)}^2+{\left(\frac{3}{\sqrt{13}}\right)}^2=\frac{4}{13}+\frac{3}{13}=\frac{13}{13}=1]$

$y={\cos }^{-1}\{\cos \alpha \sin 6-\sin \alpha \cos \theta \}$

Now $\left[\sin \right(\theta -\alpha )=\sin 6\cos \alpha -\cos 6\sin \alpha ]$

$y={\cos }^{-1}\left\{\sin (\theta -\alpha )\right\}[\cos \left(\frac{\pi }{2}(\theta -\alpha )\right)=\sin (\theta -\alpha )]$

$y={\cos }^{-1}\left\{\cos \left(\frac{\pi }{2}(\theta -\alpha )\right)\right\}$

$y=\pi /2-(\theta -\alpha )\left[{\cos }^{-1}\right(\cos x)=x]$

$y=\pi /2-({\sin }^{-1}x-{\cos }^{-1}\frac{2}{\sqrt{13}})$

Now differente wrt $x$ we have

$\frac{dy}{dx}=0-\frac{1}{\sqrt{1-x^2}}-0({\cos }^{{}_1}\frac{2}{\sqrt{13}}=\cos \tan )$

$\Rightarrow \frac{dy}{dx}=\frac{-1}{\sqrt{1-x^2}}$