Derivative of Sum of Inverse Tangent Functions

To find the derivative $\frac{dy}{dx}$ at $x = \frac{\sqrt{3}}{2}$, we can split the function into two parts, $y = y_1 + y_2$, where:

$y_1 = \tan^{-1}\left(\frac{3\cos x - 4\sin x}{4\cos x + 3\sin x}\right)$ $y_2 = 2\tan^{-1}\left(\frac{x}{1 + \sqrt{1 - x^2}}\right)$

Step 1: Simplify and Differentiate $y_1$

Divide the numerator and denominator inside the argument of $y_1$ by $4\cos x$:

$y_1 = \tan^{-1}\left(\frac{\frac{3}{4} - \tan x}{1 + \frac{3}{4}\tan x}\right)$

Let $\tan \alpha = \frac{3}{4}$, where $\alpha = \tan^{-1}\left(\frac{3}{4}\right)$ is a constant. Using the identity $\tan(\alpha - x) = \frac{\tan \alpha - \tan x}{1 + \tan \alpha \tan x}$, we have:

$y_1 = \tan^{-1}(\tan(\alpha - x)) = \alpha - x + k\pi \quad (k \in \mathbb{Z})$

Differentiating $y_1$ with respect to $x$:

$\frac{dy_1}{dx} = \frac{d}{dx}(\alpha - x) = -1$

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Step 2: Simplify and Differentiate $y_2$

Let $x = \sin \theta$, where $\theta = \sin^{-1} x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

Then $\sqrt{1 - x^2} = \cos \theta$. Substituting this into $y_2$:

$y_2 = 2\tan^{-1}\left(\frac{\sin \theta}{1 + \cos \theta}\right)$

Using half-angle trigonometric identities $\sin \theta = 2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)$ and $1 + \cos \theta = 2\cos^2\left(\frac{\theta}{2}\right)$:

$y_2 = 2\tan^{-1}\left(\frac{2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)}{2\cos^2\left(\frac{\theta}{2}\right)}\right) = 2\tan^{-1}\left(\tan\frac{\theta}{2}\right)$

Since $\frac{\theta}{2} \in \left(-\frac{\pi}{4}, \frac{\pi}{4}\right)$, we get:

$y_2 = 2 \cdot \frac{\theta}{2} = \theta = \sin^{-1} x$

Differentiating $y_2$ with respect to $x$:

$\frac{dy_2}{dx} = \frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1 - x^2}}$

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Step 3: Calculate $\frac{dy}{dx}$ at $x = \frac{\sqrt{3}}{2}$

Combining the derivatives of $y_1$ and $y_2$:

$\frac{dy}{dx} = \frac{dy_1}{dx} + \frac{dy_2}{dx} = -1 + \frac{1}{\sqrt{1 - x^2}}$

Now, evaluate at $x = \frac{\sqrt{3}}{2}$:

$\left.\frac{dy}{dx}\right|_{x = \frac{\sqrt{3}}{2}} = -1 + \frac{1}{\sqrt{1 - \left(\frac{\sqrt{3}}{2}\right)^2}} = -1 + \frac{1}{\sqrt{1 - \frac{3}{4}}} = -1 + \frac{1}{\sqrt{\frac{1}{4}}} = -1 + 2 = 1$

Thus, the required value is $1$, which corresponds to Option C.