If y= ^-1((3 x-4 x)/(4 x+3 x))+2 ^-1(x1+√1-x^2), then (dy)/(dx) at x=√32 is equal to :

Question

If $y={	an }^{-1}(\frac{3\cos x-4\sin x}{4\cos x+3\sin x})+2{	an }^{-1}(\frac{x}{1+\sqrt{1-{x}^{2}}})$, then $\frac{dy}{dx}$ at $x=\frac{\sqrt{3}}{2}$ is equal to :

Accepted Answer

$\begin{gathered} y={	an }^{-1}(\frac{\frac{3}{4}-	an x}{1+\frac{3}{4}	an x})+2{	an }^{-1}(\frac{x}{1+\sqrt{1-{x}^{2}}}) \ \cos x: \ \frac{3\cos x-4\sin x}{4\cos x+3\sin x}=\frac{3-4	an x}{4+3	an x}=\frac{\frac{3}{4}-	an x}{1+\frac{3}{4}	an x} \ {	an }^{-1}(\frac{a-b}{1+ab})={	an }^{-1}a-{	an }^{-1}b: \ {	an }^{-1}(\frac{\frac{3}{4}-	an x}{1+\frac{3}{4}	an x})={	an }^{-1}(\frac{3}{4})-{	an }^{-1}(	an x) \ \, 2{	an }^{-1}(\frac{x}{1+\sqrt{1-{x}^{2}}}) \ \, x=\sin 	heta \, \Rightarrow \, \sqrt{1-{x}^{2}}=\sqrt{1-{\sin }^{2}	heta }=\cos 	heta \ 	herefore 2{	an }^{-1}(\frac{\sin 	heta }{1+\cos 	heta })=2{	an }^{-1}(	an \frac{	heta }{2})=	heta ={\sin }^{-1}x \end{gathered}$

Hence

$\begin{gathered} y={	an }^{-1}(\frac{3}{4})-{	an }^{-1}(	an x)+{\sin }^{-1}x \ {	an }^{-1}(\frac{3}{4})0. \ \frac{d}{dx}[-{	an }^{-1}(	an x)]=-1\, ({	an }^{-1}(	an x)=x) \ \frac{d}{dx}({\sin }^{-1}x)=\frac{1}{\sqrt{1-{x}^{2}}} \ 	herefore \frac{dy}{dx}=-1+\frac{1}{\sqrt{1-{x}^{2}}} \ x=\frac{\sqrt{3}}{2}\Rightarrow \sqrt{1-{x}^{2}}=\sqrt{1-\frac{3}{4}}=\frac{1}{2} \ \frac{dy}{dx}=-1+\frac{1}{1/2}=-1+2=1 \end{gathered}$

Solutions · 1