Class CBSE Class 12 Mathematics Derivatives Q #600
COMPETENCY BASED
APPLY
1 Marks 2024 AISSCE(Board Exam) MCQ SINGLE
Derivative of \(e^{\sin^{2}x}\) with respect to cos x is:
(A) \(sin~x~e^{sin^{2}x}\)
(B) \(cos~x~e^{sin^{2}x}\)
(C) \(-2~cos~x~e^{sin^{2}x}\)
(D) \(-2~sin^{2}x~cos~x~e^{sin^{2}x}\)
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Correct Answer: C

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Detailed Solution

Step 1: Define the functions

Let $u = e^{\sin^2 x}$ and $v = \cos x$. We need to find $\frac{du}{dv}$.

Step 2: Find $\frac{du}{dx}$

Using the chain rule, we have $$ \frac{du}{dx} = \frac{d}{dx} (e^{\sin^2 x}) = e^{\sin^2 x} \cdot \frac{d}{dx} (\sin^2 x) = e^{\sin^2 x} \cdot 2 \sin x \cdot \frac{d}{dx} (\sin x) = e^{\sin^2 x} \cdot 2 \sin x \cdot \cos x = 2 \sin x \cos x e^{\sin^2 x} $$

Step 3: Find $\frac{dv}{dx}$

We have $v = \cos x$, so $$ \frac{dv}{dx} = \frac{d}{dx} (\cos x) = -\sin x $$

Step 4: Find $\frac{du}{dv}$

Using the chain rule, we have $$ \frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{2 \sin x \cos x e^{\sin^2 x}}{-\sin x} = -2 \cos x e^{\sin^2 x} $$

Final Answer: \(-2~cos~x~e^{sin^{2}x}\)

AI Suggestion: Option C

AI generated content. Review strictly for academic accuracy.

Pedagogical Audit
Bloom's Analysis: This is an APPLY question because it requires the student to apply the chain rule of differentiation to find the derivative of a composite function.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the student to apply the procedural knowledge of differentiation, specifically the chain rule, to solve the problem.<\/span>
Syllabus Audit: In the context of CBSE Class 12, this is classified as application-oriented. It tests the student's ability to apply differentiation techniques to solve a problem.
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