Class CBSE Class 12 Mathematics Derivatives Q #602
COMPETENCY BASED
APPLY
1 Marks 2024 AISSCE(Board Exam) MCQ SINGLE
Derivative of \(e^{2x}\) with respect to \(e^{x}\), is:
(A) \(e^{x}\)
(B) \(2e^{x}\)
(C) \(2e^{2x}\)
(D) \(2e^{3x}\)
Prev Next

AI Tutor Explanation

Powered by Gemini

Step-by-Step Solution

Let \(u = e^{2x}\) and \(v = e^{x}\). We want to find \(\frac{du}{dv}\).
First, find \(\frac{du}{dx}\) and \(\frac{dv}{dx}\): \[ \frac{du}{dx} = \frac{d}{dx}(e^{2x}) = 2e^{2x} \] \[ \frac{dv}{dx} = \frac{d}{dx}(e^{x}) = e^{x} \]
Now, use the chain rule: \[ \frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{2e^{2x}}{e^{x}} = 2e^{2x-x} = 2e^{x} \]

Correct Answer: 2e^{x}

AI Suggestion: Option B

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 and differentiation formulas to find the derivative of one function with respect to another.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the student to execute a sequence of steps involving differentiation and the chain rule, which falls under the procedural knowledge dimension.
Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. It assesses the student's ability to apply differentiation techniques in a slightly non-standard way, requiring them to find the derivative of one function with respect to another, rather than with respect to \(x\). This goes beyond rote memorization and tests conceptual understanding and application.

More from this Chapter

SA
If $x=e^{cos~3t}$ and $y=e^{sin~3t}$ , prove that $\frac{dy}{dx}=-\frac{y~log~x}{x~log~y}$
SA
Differentiate $y=\sin^{-1}(3x-4x^{3})$ w.r.t. x, if $x\in[-\frac{1}{2},\frac{1}{2}]$.
MCQ_SINGLE
The derivative of \(\sin(x^{2})\) w.r.t. x, at \(x=\sqrt{\pi}\) is :
VSA
Differentiate $\left(\frac{5^{x}}{x^{5}}\right)$ with respect to x.
MCQ_SINGLE
Derivative of \(e^{\sin^{2}x}\) with respect to cos x is:
View All Questions