Question

Find \( \frac{dy}{dx} \), if \( y = \tan^{-1} \left( \frac{2x}{1 - x^2} \right) \)

Hint:

To differentiate \( \tan^{-1} \left( \frac{u}{v} \right) \), use the chain rule and quotient rule as shown.

Answer 1

Step 1: Differentiating the inverse tangent function.
We are given \( y = \tan^{-1} \left( \frac{2x}{1 - x^2} \right) \). Using the chain rule, we differentiate the inverse tangent function. The derivative of \( \tan^{-1}(u) \) with respect to \( u \) is \( \frac{1}{1 + u^2} \), so: \[ \frac{dy}{dx} = \frac{1}{1 + \left( \frac{2x}{1 - x^2} \right)^2} \cdot \frac{d}{dx} \left( \frac{2x}{1 - x^2} \right) \] Step 2: Differentiating \( \frac{2x}{1 - x^2} \).
We apply the quotient rule to differentiate \( \frac{2x}{1 - x^2} \). The quotient rule states that if \( u = 2x \) and \( v = 1 - x^2 \), then: \[ \frac{d}{dx} \left( \frac{2x}{1 - x^2} \right) = \frac{(1 - x^2)(2) - (2x)(-2x)}{(1 - x^2)^2} \] Simplifying: \[ \frac{d}{dx} \left( \frac{2x}{1 - x^2} \right) = \frac{2(1 - x^2) + 4x^2}{(1 - x^2)^2} = \frac{2 + 2x^2}{(1 - x^2)^2} \] Step 3: Final expression for \( \frac{dy}{dx} \).
Substituting this into the derivative of \( y \), we get: \[ \frac{dy}{dx} = \frac{1}{1 + \left( \frac{2x}{1 - x^2} \right)^2} \cdot \frac{2 + 2x^2}{(1 - x^2)^2} \] Simplifying the denominator \( 1 + \left( \frac{2x}{1 - x^2} \right)^2 \): \[ 1 + \left( \frac{2x}{1 - x^2} \right)^2 = \frac{(1 - x^2)^2 + 4x^2}{(1 - x^2)^2} = \frac{1 - 2x^2 + x^4 + 4x^2}{(1 - x^2)^2} = \frac{1 + 2x^2 + x^4}{(1 - x^2)^2} \] Thus, \[ \frac{dy}{dx} = \frac{2 + 2x^2}{1 + 2x^2 + x^4} \] Step 4: Conclusion.
Therefore, the derivative of \( y \) is: \[ \frac{dy}{dx} = \frac{2(1 + x^2)}{(1 + x^2)^2} = \frac{2}{1 + x^2} \]