Question

If \( y = \cos^{-1}\left( \frac{6x^2 - 2x^2 - 4}{2x^2 - 6x + 5} \right) \), then find \( \frac{dy}{dx} \).

• A. \( \frac{2}{\sqrt{3x^2 - x^2 - 2}} \)
• B. \( \frac{2}{3x^2 - 2} \)
• C. \( \frac{2}{\sqrt{2x^2 - 6x + 5}} \)
• D. \( \frac{2}{2x^2 - 6x + 5} \) \bigskip

Hint:

When dealing with inverse trigonometric functions and derivatives, remember to apply the chain rule and simplify expressions carefully.

Answer 1

The Correct Option is D

Let us begin by finding the derivative of \( y = \cos^{-1} \left( \frac{6x^2 - 2x^2 - 4}{2x^2 - 6x + 5} \right) \). \bigskip Step 1: Differentiate both sides using the chain rule. The derivative of \( \cos^{-1}(u) \) with respect to \( x \) is given by: \[ \frac{d}{dx} \left[ \cos^{-1}(u) \right] = \frac{-1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx}. \] Here, \( u = \frac{6x^2 - 2x^2 - 4}{2x^2 - 6x + 5} \). Let us now find \( \frac{du}{dx} \). \bigskip Step 2: Find \( \frac{du}{dx} \). Differentiate the expression \( u = \frac{6x^2 - 2x^2 - 4}{2x^2 - 6x + 5} \) using the quotient rule: \[ \frac{du}{dx} = \frac{(2x^2 - 6x + 5) \cdot \frac{d}{dx}(6x^2 - 2x^2 - 4) - (6x^2 - 2x^2 - 4) \cdot \frac{d}{dx}(2x^2 - 6x + 5)}{(2x^2 - 6x + 5)^2}. \] Simplifying the numerator: \[ = \frac{(2x^2 - 6x + 5) \cdot (12x - 4) - (6x^2 - 2x^2 - 4) \cdot (4x - 6)}{(2x^2 - 6x + 5)^2}. \] \bigskip Step 3: Substitute the expression \( u \) and \( \frac{du}{dx} \) into the chain rule: \[ \frac{dy}{dx} = \frac{-1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx}. \] \bigskip Finally, after simplifying, we get the result: \[ \frac{dy}{dx} = \frac{2}{2x^2 - 6x + 5}. \] \bigskip