Question

Let $y = \tan^{-1}\left( \frac{3\cos x - 4\sin x}{4\cos x + 3\sin x} \right) + \tan^{-1}\left( \frac{x}{1 + \sqrt{1 + x^2}} \right)$. Then the value of $\frac{dy}{dx}$ at $x = \frac{\sqrt{3}}{2}$ is :

• A. $-\frac{5}{7}$
• B. $\frac{3}{7}$
• C. $-\frac{3}{2}$
• D. $\frac{5}{7}$

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The problem involves differentiating a sum of two inverse trigonometric functions.
To simplify the expression, we use trigonometric identities like $\tan^{-1}\left( \frac{a - b}{1 + ab} \right) = \tan^{-1}a - \tan^{-1}b$ and half-angle formulas.
Step 2: Key Formula or Approach:
1. For the first term, divide the numerator and denominator by $4\cos x$ to reach the standard $\tan^{-1}$ identity form.
2. For the second term, use the substitution $x = \tan \theta$ to simplify the square root and the fraction.
Step 3: Detailed Explanation:
1. Simplifying the first term, $y_1$:
\[ y_1 = \tan^{-1}\left( \frac{\frac{3\cos x}{4\cos x} - \frac{4\sin x}{4\cos x}}{\frac{4\cos x}{4\cos x} + \frac{3\sin x}{4\cos x}} \right) = \tan^{-1}\left( \frac{\frac{3}{4} - \tan x}{1 + \frac{3}{4}\tan x} \right) \]
Using $\tan^{-1}\left( \frac{a - b}{1 + ab} \right) = \tan^{-1}a - \tan^{-1}b$:
\[ y_1 = \tan^{-1}\left(\frac{3}{4}\right) - \tan^{-1}(\tan x) = \tan^{-1}\left(\frac{3}{4}\right) - x \]
2. Simplifying the second term, $y_2$:
Let $x = \tan \theta \implies \theta = \tan^{-1}x$.
\[ y_2 = \tan^{-1}\left( \frac{\tan \theta}{1 + \sqrt{1 + \tan^2 \theta}} \right) = \tan^{-1}\left( \frac{\tan \theta}{1 + \sec \theta} \right) \]
\[ y_2 = \tan^{-1}\left( \frac{\sin \theta / \cos \theta}{(1 + \cos \theta) / \cos \theta} \right) = \tan^{-1}\left( \frac{2\sin(\theta/2)\cos(\theta/2)}{2\cos^2(\theta/2)} \right) = \tan^{-1}(\tan(\theta/2)) = \frac{\theta}{2} \]
Substituting $\theta = \tan^{-1}x$:
\[ y_2 = \frac{1}{2}\tan^{-1}x \]
3. Combining and differentiating:
\[ y = \tan^{-1}\left(\frac{3}{4}\right) - x + \frac{1}{2}\tan^{-1}x \]
\[ \frac{dy}{dx} = 0 - 1 + \frac{1}{2(1 + x^2)} \]
4. Substituting $x = \frac{\sqrt{3}}{2} \implies x^2 = \frac{3}{4}$:
\[ \left. \frac{dy}{dx} \right|_{x=\frac{\sqrt{3}}{2}} = -1 + \frac{1}{2\left( 1 + \frac{3}{4} \right)} = -1 + \frac{1}{2 \cdot \frac{7}{4}} = -1 + \frac{2}{7} = -\frac{5}{7} \]
Step 4: Final Answer:
The derivative at $x = \frac{\sqrt{3}}{2}$ is $-\frac{5}{7}$.