Question

If \( y = \sin^{-1}(3x - 4x^3) \), find \( \dfrac{dy}{dx}. \)

• A. \( \dfrac{3 - 12x^2}{\sqrt{1-(3x-4x^3)^2}} \)
• B. \( \dfrac{3 - 12x^2}{\sqrt{1-(3x-4x^3)}} \)
• C. \( \dfrac{12x^2 - 3}{\sqrt{1-(3x-4x^3)^2}} \)
• D. \( \dfrac{3}{\sqrt{1-(3x-4x^3)^2}} \)

Hint:

When differentiating inverse trigonometric functions, first set the inner expression as \(u\), then apply \( \frac{d}{dx}(\sin^{-1}u) = \frac{u'}{\sqrt{1-u^2}} \).

Answer 1

The Correct Option is A

Concept: The derivative of the inverse sine function is \[ \frac{d}{dx}(\sin^{-1}u) = \frac{u'}{\sqrt{1-u^2}} \] where \(u\) is a function of \(x\).

Step 1: Identify \(u\). \[ y = \sin^{-1}(3x - 4x^3) \] Let \[ u = 3x - 4x^3 \]

Step 2: Differentiate \(u\). \[ \frac{du}{dx} = 3 - 12x^2 \]

Step 3: Apply the derivative formula. \[ \frac{dy}{dx} = \frac{3 - 12x^2}{\sqrt{1-(3x-4x^3)^2}} \]