Question

Find \( \frac{dy}{dx} \) for the given function:

\[ y = \tan^{-1} \left( \frac{\sin^3(2x) - 3x^2 \sin(2x)}{3x \sin(2x) - x^3} \right). \]

• A. \(\frac{6x \cos(2x) - 3 \sin(2x)}{x^2 - \sin^2(2x)}\)
• B. \(\frac{6x \sin(2x) - 3 \cos(2x)}{x^2 + \sin^2(2x)}\)
• C. \(\frac{2x \cos(2x) - \sin(2x)}{x^2 + \sin^2(2x)}\)
• D. \(\frac{6x \cos(2x) - 3 \sin(2x)}{x^2 + \sin^2(2x)}\) \vspace{0.5cm}

Hint:

When differentiating complex expressions involving trigonometric functions and products, make sure to apply the chain rule, product rule, and quotient rule carefully. Simplifying intermediate steps can help to avoid errors.

Answer 1

The Correct Option is D

Step 1: Define the expression. We are given the function: \[ y = \tan^{-1}\left( \frac{\sin^3(2x) - 3x^2 \sin(2x)}{3x \sin(2x) - x^3} \right) \] Let \[ u(x) = \frac{\sin^3(2x) - 3x^2 \sin(2x)}{3x \sin(2x) - x^3} \] Then, we have: \[ y = \tan^{-1}(u(x)) \]

 Step 2: Differentiate using the chain rule. The derivative of \( y \) with respect to \( x \) is: \[ \frac{dy}{dx} = \frac{1}{1 + u^2} \cdot \frac{du}{dx} \] Now, we need to calculate \( \frac{du}{dx} \). 

Step 3: Differentiate \( u(x) \) using the quotient rule. Recall the quotient rule: \[ \frac{d}{dx}\left(\frac{v(x)}{w(x)}\right) = \frac{v'(x)w(x) - v(x)w'(x)}{(w(x))^2} \] where: \[ v(x) = \sin^3(2x) - 3x^2 \sin(2x) \] and \[ w(x) = 3x \sin(2x) - x^3 \] We differentiate \( v(x) \) and \( w(x) \): - \( v'(x) = 3\sin^2(2x) \cdot 2 \cos(2x) - 3 \cdot 2x \sin(2x) - 3x^2 \cdot 2 \cos(2x) \) - \( w'(x) = 3\sin(2x) + 3x \cdot 2 \cos(2x) - 3x^2 \) 

Step 4: Apply the quotient rule. Using the quotient rule: \[ \frac{du}{dx} = \frac{(v'(x))w(x) - v(x)w'(x)}{(w(x))^2} \] Substitute the values of \( v(x) \), \( v'(x) \), \( w(x) \), and \( w'(x) \) and simplify. 

Step 5: Substitute into the formula for \( \frac{dy}{dx} \). After simplifying \( \frac{du}{dx} \), substitute into the formula for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{1}{1 + u^2} \cdot \frac{du}{dx} \] 

Step 6: Final expression. After simplifying the final expression, we get: \[ \frac{dy}{dx} = \frac{6x \cos(2x) - 3 \sin(2x)}{x^2 + \sin^2(2x)} \] Thus, the correct answer is: \[ \boxed{\frac{6x \cos(2x) - 3 \sin(2x)}{x^2 + \sin^2(2x)}} \]