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Question: find derivative of $\sin^{-1}(3x-4x^3)$ for different intervals of x...

find derivative of sin⁡−1(3x−4x3)\sin^{-1}(3x-4x^3) for different intervals of x

Answer

The derivative of sin⁡−1(3x−4x3)\sin^{-1}(3x-4x^3) is:

ddx(sin⁡−1(3x−4x3))={31−x2if −12<x<12−31−x2if −1≤x<−12 or 12<x≤1\frac{d}{dx}(\sin^{-1}(3x-4x^3)) = \begin{cases} \frac{3}{\sqrt{1-x^2}} & \text{if } -\frac{1}{2} < x < \frac{1}{2} \\ -\frac{3}{\sqrt{1-x^2}} & \text{if } -1 \le x < -\frac{1}{2} \text{ or } \frac{1}{2} < x \le 1 \end{cases}
Explanation

Solution

Let x=sin⁡θx = \sin\theta. Then 3x−4x3=sin⁡(3θ)3x-4x^3 = \sin(3\theta). The function becomes sin⁡−1(sin⁡(3θ))\sin^{-1}(\sin(3\theta)). The identity sin⁡−1(sin⁡y)\sin^{-1}(\sin y) depends on the interval of yy.

  1. If −π2≤3θ≤π2-\frac{\pi}{2} \le 3\theta \le \frac{\pi}{2}, which means −12≤x≤12-\frac{1}{2} \le x \le \frac{1}{2}, then sin⁡−1(sin⁡(3θ))=3θ=3sin⁡−1x\sin^{-1}(\sin(3\theta)) = 3\theta = 3\sin^{-1}x. Derivative is 31−x2\frac{3}{\sqrt{1-x^2}}.
  2. If π2<3θ≤3π2\frac{\pi}{2} < 3\theta \le \frac{3\pi}{2}, which means 12<x≤1\frac{1}{2} < x \le 1, then sin⁡−1(sin⁡(3θ))=π−3θ=π−3sin⁡−1x\sin^{-1}(\sin(3\theta)) = \pi - 3\theta = \pi - 3\sin^{-1}x. Derivative is −31−x2-\frac{3}{\sqrt{1-x^2}}.
  3. If −3π2≤3θ<−π2-\frac{3\pi}{2} \le 3\theta < -\frac{\pi}{2}, which means −1≤x<−12-1 \le x < -\frac{1}{2}, then sin⁡−1(sin⁡(3θ))=−π−3θ=−π−3sin⁡−1x\sin^{-1}(\sin(3\theta)) = -\pi - 3\theta = -\pi - 3\sin^{-1}x. Derivative is −31−x2-\frac{3}{\sqrt{1-x^2}}.

The derivative is not defined at x=±1x = \pm 1 because 1−x2\sqrt{1-x^2} becomes zero. The function is also not differentiable at x=±12x = \pm \frac{1}{2} as the left and right derivatives do not match.

Combining the results for the open intervals:

  • For x∈(−12,12)x \in (-\frac{1}{2}, \frac{1}{2}), the derivative is 31−x2\frac{3}{\sqrt{1-x^2}}.
  • For x∈[−1,−12)∪(12,1]x \in [-1, -\frac{1}{2}) \cup (\frac{1}{2}, 1], the derivative is −31−x2-\frac{3}{\sqrt{1-x^2}}.