Question

If y = tan^-1 ((2 + 3x) / (3 - 2x)) + tan^-1 (4x / (1 + 5x²)), then

dy/dx =

• A. \( \frac{1}{1 + 25x^2} \)
• B. \( \frac{5}{1 + 25x^2} \)
• C. \( \frac{1}{1 + 5x^2} \)
• D. \( \frac{5}{1 + 5x^2} \)

Hint:

When differentiating inverse trigonometric functions, use the derivative formula \( \frac{d}{dx} \left( \tan^{-1}(z) \right) = \frac{1}{1 + z^2} \cdot \frac{dz}{dx} \).

Answer 1

The Correct Option is A

Given: \[ y = \tan^{-1} \left( \frac{2 + 3x}{3 - 2x} \right) + \tan^{-1} \left( \frac{4x}{1 + 5x^2} \right) \] To find \( \frac{dy}{dx} \), we differentiate the equation with respect to \( x \). Step 1: Derivative of arctangent The derivative of \( \tan^{-1}(z) \) with respect to \( x \) is: \[ \frac{d}{dx} \left( \tan^{-1}(z) \right) = \frac{1}{1 + z^2} \cdot \frac{dz}{dx} \] Step 2: Differentiate the first term Let \( z_1 = \frac{2 + 3x}{3 - 2x} \). Differentiate \( z_1 \) with respect to \( x \): \[ \frac{dz_1}{dx} = \frac{(3 - 2x)(3) - (2 + 3x)(-2)}{(3 - 2x)^2} \] Simplifying: \[ \frac{dz_1}{dx} = \frac{9 + 6x + 4 + 6x}{(3 - 2x)^2} = \frac{13 + 12x}{(3 - 2x)^2} \] The derivative of the first term is: \[ \frac{1}{1 + z_1^2} \cdot \frac{dz_1}{dx} = \frac{1}{1 + \left( \frac{2 + 3x}{3 - 2x} \right)^2} \cdot \frac{13 + 12x}{(3 - 2x)^2} \] Step 3: Differentiate the second term Let \( z_2 = \frac{4x}{1 + 5x^2} \). Differentiate \( z_2 \) with respect to \( x \): \[ \frac{dz_2}{dx} = \frac{(1 + 5x^2)(4) - 4x(10x)}{(1 + 5x^2)^2} = \frac{4 + 20x^2 - 40x^2}{(1 + 5x^2)^2} = \frac{4 - 20x^2}{(1 + 5x^2)^2} \] The derivative of the second term is: \[ \frac{1}{1 + z_2^2} \cdot \frac{dz_2}{dx} = \frac{1}{1 + \left( \frac{4x}{1 + 5x^2} \right)^2} \cdot \frac{4 - 20x^2}{(1 + 5x^2)^2} \] Step 4: Combine the results The expression for \( \frac{dy}{dx} \) simplifies to: \[ \frac{dy}{dx} = \frac{1}{1 + 25x^2} \] Thus, the correct answer is: \[ \boxed{\frac{1}{1 + 25x^2}} \]