If y = tan^{-1} left( frac{3cos x - 4sin x}{4 | vedclass

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(C) Let $y = y_1 + y_2$, where $y_1 = 	an^{-1} \left( \frac{3\cos x - 4\sin x}{4\cos x + 3\sin x} ight)$ and $y_2 = 2	an^{-1} \left( \frac{x}{1+\s

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(C) Let $y = y_1 + y_2$, where $y_1 = 	an^{-1} \left( \frac{3\cos x - 4\sin x}{4\cos x + 3\sin x} ight)$ and $y_2 = 2	an^{-1} \left( \frac{x}{1+\sqrt{1-x^2}} ight)$.For $y_1$, divide numerator and denominator by $4\cos x$: $y_1 = 	an^{-1} \left( \frac{3/4 - 	an x}{1 + (3/4)	an x} ight) = 	an^{-1}(3/4) - x$.Thus, $\frac{dy_1}{dx} = -1$.For $y_2$, let $x = \sin 	heta$, then $	heta = \sin^{-1} x$. The expression becomes $2	an^{-1} \left( \frac{\sin 	heta}{1+\cos 	heta} ight) = 2	an^{-1} \left( 	an(	heta/2) ight) = 	heta = \sin^{-1} x$.Thus, $\frac{dy_2}{dx} = \frac{1}{\sqrt{1-x^2}}$.Therefore, $\frac{dy}{dx} = -1 + \frac{1}{\sqrt{1-x^2}}$.At $x = \frac{\sqrt{3}}{2}$, $\frac{dy}{dx} = -1 + \frac{1}{\sqrt{1 - 3/4}} = -1 + \frac{1}{\sqrt{1/4}} = -1 + 2 = 1$.

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

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