Question

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

• A. $5/7$
• B. $-5/7$
• C. $3/7$
• D. $-3/2$

Hint:

Always simplify expressions inside $\tan^{-1}$ using trigonometric identities or substitutions before calculating the derivative. It converts a complex differentiation task into a simple one.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The function $y$ consists of two inverse trigonometric terms. We should simplify each term using substitutions before differentiating.
Step 2: Key Formula or Approach:
1. $\tan^{-1} \left( \frac{a - b}{1 + ab} \right) = \tan^{-1} a - \tan^{-1} b$
2. Half-angle substitutions for $\tan^{-1}$.
Step 3: Detailed Explanation:
Simplifying the first term ($T_1$):
Divide numerator and denominator by $4 \cos x$:
\[ T_1 = \tan^{-1} \left( \frac{\frac{3}{4} - \tan x}{1 + \frac{3}{4} \tan x} \right) \]
Using the identity, let $\tan \alpha = 3/4$:
\[ T_1 = \tan^{-1}(\tan \alpha) - \tan^{-1}(\tan x) = \alpha - x \]
Simplifying the second term ($T_2$):
Let $x = \tan \theta$, then $\sqrt{1+x^2} = \sec \theta$:
\[ T_2 = \tan^{-1} \left( \frac{\tan \theta}{1 + \sec \theta} \right) = \tan^{-1} \left( \frac{\sin \theta / \cos \theta}{(1 + \cos \theta) / \cos \theta} \right) \]
\[ T_2 = \tan^{-1} \left( \frac{\sin \theta}{1 + \cos \theta} \right) = \tan^{-1} \left( \frac{2 \sin(\theta/2) \cos(\theta/2)}{2 \cos^2(\theta/2)} \right) \]
\[ T_2 = \tan^{-1}(\tan(\theta/2)) = \frac{\theta}{2} = \frac{1}{2} \tan^{-1} x \]
The complete function is:
\[ y = \alpha - x + \frac{1}{2} \tan^{-1} x \]
Differentiating with respect to $x$:
\[ \frac{dy}{dx} = 0 - 1 + \frac{1}{2(1 + x^2)} \]
At $x = \frac{\sqrt{3}}{2}$, $x^2 = \frac{3}{4}$:
\[ \frac{dy}{dx} = -1 + \frac{1}{2(1 + \frac{3}{4})} = -1 + \frac{1}{2(\frac{7}{4})} = -1 + \frac{2}{7} = -\frac{5}{7} \]
Step 4: Final Answer:
The derivative at $x = \frac{\sqrt{3}}{2}$ is $-5/7$.