Question

If \( y = (\tan^{-1} x)^2 \), prove that \( (x^2 + 1)^2 y_2 + 2x (x^2 + 1) y_1 = 2 \).

Hint:

When working with inverse trigonometric functions, use the chain rule and product rule for differentiation. For higher derivatives, apply the product and quotient rules carefully.

Answer 1

Step 1: Find the first and second derivatives of \( y \).
Given \( y = (\tan^{-1} x)^2 \), we differentiate \( y \) with respect to \( x \). Using the chain rule: \[ \frac{dy}{dx} = 2 \tan^{-1}(x) \cdot \frac{d}{dx} \left( \tan^{-1}(x) \right) \] Since \( \frac{d}{dx} \left( \tan^{-1}(x) \right) = \frac{1}{1 + x^2} \), we get: \[ \frac{dy}{dx} = 2 \tan^{-1}(x) \cdot \frac{1}{1 + x^2} \] Thus, \( y_1 = 2 \tan^{-1}(x) \cdot \frac{1}{1 + x^2} \). Step 2: Find the second derivative of \( y \).
To find \( y_2 \), we differentiate \( y_1 \) again using the product rule: \[ y_2 = \frac{d}{dx} \left[ 2 \tan^{-1}(x) \cdot \frac{1}{1 + x^2} \right] \] The product rule gives: \[ y_2 = 2 \cdot \frac{1}{1 + x^2} \cdot \frac{d}{dx} \left( \tan^{-1}(x) \right) + 2 \tan^{-1}(x) \cdot \frac{d}{dx} \left( \frac{1}{1 + x^2} \right) \] Simplifying: \[ y_2 = 2 \cdot \frac{1}{1 + x^2} \cdot \frac{1}{1 + x^2} - 2 \tan^{-1}(x) \cdot \frac{2x}{(1 + x^2)^2} \] Step 3: Proving the equation.
Using the derivatives \( y_1 \) and \( y_2 \), we can now prove the given equation: \[ (x^2 + 1)^2 y_2 + 2x (x^2 + 1) y_1 = 2 \]