Class CBSE Class 12 Mathematics Derivatives Q #1298
KNOWLEDGE BASED
APPLY
3 Marks 2024 AISSCE(Board Exam) SA
If $y=(tan~x)^{x}$, then find $\frac{dy}{dx}$ .
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AI Tutor Explanation

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Detailed Solution

Step 1: Apply Logarithmic Transformation

Given $y = (\tan x)^x$. To find $\frac{dy}{dx}$, we first take the natural logarithm of both sides:

Step 2: Differentiate both sides with respect to x

Differentiating both sides with respect to $x$, we get: $$\frac{1}{y} \frac{dy}{dx} = x \cdot \frac{1}{\tan x} \cdot \sec^2 x + \ln(\tan x) \cdot 1$$

Step 3: Simplify the expression

Simplify the expression: $$\frac{1}{y} \frac{dy}{dx} = \frac{x \sec^2 x}{\tan x} + \ln(\tan x)$$ $$\frac{1}{y} \frac{dy}{dx} = \frac{x}{\sin x \cos x} + \ln(\tan x)$$

Step 4: Solve for dy/dx

Multiply both sides by $y$: $$\frac{dy}{dx} = y \left( \frac{x}{\sin x \cos x} + \ln(\tan x) \right)$$ Substitute $y = (\tan x)^x$: $$\frac{dy}{dx} = (\tan x)^x \left( \frac{x}{\sin x \cos x} + \ln(\tan x) \right)$$ $$\frac{dy}{dx} = (\tan x)^x \left( 2x \csc(2x) + \ln(\tan x) \right)$$

Final Answer: $\frac{dy}{dx} = (\tan x)^x \left( \frac{x}{\sin x \cos x} + \ln(\tan x) \right)$

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because the student needs to apply the rules of logarithmic differentiation to solve the problem.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the student to apply a specific procedure (logarithmic differentiation) to find the derivative.<\/span>
Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. The question directly tests the student's knowledge of differentiation techniques, specifically logarithmic differentiation, which is a standard topic in the syllabus.

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