Question

If \( y = \sin^{-1}(3x - 4x^3) \), find the derivative \( \dfrac{dy}{dx} \) in its standard form.

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

Hint:

For inverse trigonometric functions like \( \sin^{-1}(u) \), always remember to apply the chain rule: \[ \frac{d}{dx}(\sin^{-1}u)=\frac{u'}{\sqrt{1-u^2}} \] where \(u\) is a function of \(x\).

Answer 1

The Correct Option is A

Concept: The derivative of the inverse sine function is: \[ \frac{d}{dx}\left(\sin^{-1} u\right)=\frac{1}{\sqrt{1-u^2}}\cdot\frac{du}{dx} \] Thus, when differentiating \( \sin^{-1}(u) \), we apply the chain rule: \[ \frac{dy}{dx}=\frac{1}{\sqrt{1-u^2}}\cdot\frac{du}{dx} \]

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

Step 2: Differentiate the inner function. \[ \frac{du}{dx} = \frac{d}{dx}(3x-4x^3) \] \[ = 3 - 12x^2 \]

Step 3: Apply the chain rule. Using \[ \frac{d}{dx}(\sin^{-1}u)=\frac{1}{\sqrt{1-u^2}}\frac{du}{dx} \] \[ \frac{dy}{dx} = \frac{1}{\sqrt{1-(3x-4x^3)^2}}\,(3-12x^2) \]

Step 4: Write the derivative in standard form. \[ \boxed{\frac{dy}{dx}=\frac{3-12x^2}{\sqrt{1-(3x-4x^3)^2}}} \]