Question

If \[ \tan u = \frac{\sqrt{1 - x}}{\sqrt{1 + x}}, \quad \cos v = 4x^3 - 3x, \quad \text{then} \quad \frac{du}{dv} = \text{?}

• A. \( \frac{1}{6} \)
• B. 1
• C. 2
• D. \( \frac{1}{2} \)

Hint:

When finding \( \frac{du}{dv} \), differentiate both expressions with respect to \( x \) and apply the chain rule.

Answer 1

The Correct Option is A

Step 1: Differentiate the expressions.
We are given two functions, \( \tan u \) and \( \cos v \). First, differentiate \( \tan u \) with respect to \( x \): \[ \frac{d}{dx} \left( \tan u \right) = \sec^2 u \cdot \frac{du}{dx} \] Next, differentiate \( \cos v \) with respect to \( x \): \[ \frac{d}{dx} \left( \cos v \right) = -\sin v \cdot \frac{dv}{dx} \]
Step 2: Find \( \frac{du}{dv} \).
Now, we apply the chain rule to express \( \frac{du}{dv} \) in terms of \( x \) and simplify it. We ultimately find: \[ \frac{du}{dv} = \frac{1}{6} \]
Step 3: Conclusion.
Thus, \( \frac{du}{dv} = \frac{1}{6} \), corresponding to option (A).