Question

If \( u = \tan^{-1}\!\left(\dfrac{1+x^2-1}{x}\right) \) and \( v = \tan^{-1}\!\left(\dfrac{2x(1-x^2)}{1-2x^2}\right) \), then \( \dfrac{du}{dv} \) at \( x = 0 \) is

• A. \( \dfrac{1}{4} \)
• B. \( \dfrac{1}{8} \)
• C. \( 1 \)
• D. \( -\dfrac{1}{8} \)

Hint:

To find \( \dfrac{du}{dv} \), always use \( \dfrac{du}{dv} = \dfrac{du/dx}{dv/dx} \).

Answer 1

The Correct Option is A

Step 1: Simplify \( u \).
\[ u = \tan^{-1}\left(\frac{x^2}{x}\right) = \tan^{-1}(x) \] Step 2: Differentiate \( u \) with respect to \( x \).
\[ \frac{du}{dx} = \frac{1}{1+x^2} \Rightarrow \left.\frac{du}{dx}\right|_{x=0} = 1 \] Step 3: Differentiate \( v \) with respect to \( x \).
Let \[ v = \tan^{-1}(f(x)), \quad f(x) = \frac{2x(1-x^2)}{1-2x^2} \] \[ \frac{dv}{dx} = \frac{f'(x)}{1+f(x)^2} \] At \( x = 0 \), \[ f(0)=0, \quad f'(0)=4 \Rightarrow \left.\frac{dv}{dx}\right|_{x=0} = 4 \] Step 4: Find \( \dfrac{du}{dv} \).
\[ \frac{du}{dv} = \frac{\frac{du}{dx}}{\frac{dv}{dx}} = \frac{1}{4} \] Step 5: Conclusion.
\[ \left.\frac{du}{dv}\right|_{x=0} = \frac{1}{4} \]