Question:medium

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

Show 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 \(x\).
Updated On: Apr 16, 2026
  • \( \dfrac{3-12x^2}{\sqrt{1-(3x-4x^3)^2}} \)
  • \( \dfrac{3+12x^2}{\sqrt{1-(3x-4x^3)^2}} \)
  • \( \dfrac{3-12x^2}{1-(3x-4x^3)^2} \)
  • \( \dfrac{12x^2-3}{\sqrt{1-(3x-4x^3)^2}} \)
Show Solution

The Correct Option is A

Solution and Explanation

Step 1: Understanding the Question:
The question requires us to find the derivative of \(y\) with respect to \(x\), where \(y\) is a composite function involving the inverse sine function.
Step 2: Key Formula or Approach:
The problem can be solved using the chain rule for differentiation. The formula for the derivative of the inverse sine function is:
\[ \frac{d}{dx}\left(\sin^{-1}(u)\right) = \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx} \] where \(u\) is a function of \(x\).
Step 3: Detailed Explanation:
In this problem, the function is \( y = \sin^{-1}(3x - 4x^3) \). Let's identify the inner function \(u\):
\[ u = 3x - 4x^3 \] First, find the derivative of \(u\) with respect to \(x\):
\[ \frac{du}{dx} = \frac{d}{dx}(3x - 4x^3) = 3 - 12x^2 \] Now, apply the chain rule formula:
\[ \frac{dy}{dx} = \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx} \] Substitute the expressions for \(u\) and \(\frac{du}{dx}\):
\[ \frac{dy}{dx} = \frac{1}{\sqrt{1-(3x-4x^3)^2}} \cdot (3-12x^2) \] Step 4: Final Answer:
Combining the terms into a single fraction gives the final derivative in the required form:
\[ \frac{dy}{dx} = \frac{3-12x^2}{\sqrt{1-(3x-4x^3)^2}} \] This matches option (A).
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