Class CBSE Class 12 Mathematics Derivatives Q #1451
KNOWLEDGE BASED
UNDERSTAND
3 Marks 2025 AISSCE(Board Exam) SA
Differentiate $y=\sin^{-1}(3x-4x^{3})$ w.r.t. x, if $x\in[-\frac{1}{2},\frac{1}{2}]$.
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Detailed Solution

Step 1: Trigonometric Substitution

Let $x = \sin\theta$. Then $\theta = \sin^{-1}x$. Since $x \in [-\frac{1}{2}, \frac{1}{2}]$, $\theta \in [-\frac{\pi}{6}, \frac{\pi}{6}]$.

Step 2: Simplify the expression inside the arcsine

Substitute $x = \sin\theta$ into the expression $3x - 4x^3$:\r\n$$3x - 4x^3 = 3\sin\theta - 4\sin^3\theta = \sin(3\theta)$$\r\nTherefore, $y = \sin^{-1}(\sin(3\theta))$.

Step 3: Simplify the arcsine expression

Since $\theta \in [-\frac{\pi}{6}, \frac{\pi}{6}]$, we have $3\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$. Therefore, $\sin^{-1}(\sin(3\theta)) = 3\theta$.\r\nSo, $y = 3\theta = 3\sin^{-1}x$.

Step 4: Differentiate with respect to x

Now, differentiate $y$ with respect to $x$:\r\n$$\frac{dy}{dx} = \frac{d}{dx}(3\sin^{-1}x) = 3\frac{d}{dx}(\sin^{-1}x) = 3\cdot\frac{1}{\sqrt{1-x^2}}$$\r\nThus, $\frac{dy}{dx} = \frac{3}{\sqrt{1-x^2}}$.

Final Answer: $\frac{3}{\sqrt{1-x^2}}$

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Pedagogical Audit
Bloom's Analysis: This is an UNDERSTAND question because the student needs to understand the trigonometric substitution and the range of inverse trigonometric functions to simplify the given expression before differentiating.
Knowledge Dimension: CONCEPTUAL
Justification: The question requires understanding of the concepts of inverse trigonometric functions, trigonometric identities, and differentiation. It's not just about recalling a formula but applying the concepts to solve the problem.
Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. The question directly tests the student's knowledge of inverse trigonometric functions and differentiation, which are core topics in the syllabus.

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