Class CBSE Class 12 Mathematics Derivatives Q #1313
KNOWLEDGE BASED
UNDERSTAND
2 Marks 2024 AISSCE(Board Exam) VSA
If $x^{y}=e^{x-y},$ prove that $\frac{dy}{dx}=\frac{log~x}{(1+log~x)^{2}}.$
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Detailed Solution

Step 1: Apply Logarithm on both sides

Given the equation $x^{y}=e^{x-y}$. Taking the natural logarithm (ln) on both sides, we get: $$ln(x^{y}) = ln(e^{x-y})$$ Using the logarithm property $ln(a^b) = b \cdot ln(a)$, we have: $$y \cdot ln(x) = (x-y) \cdot ln(e)$$ Since $ln(e) = 1$, the equation simplifies to: $$y \cdot ln(x) = x - y$$

Step 2: Express y in terms of x

Rearrange the equation to isolate $y$: $$y \cdot ln(x) + y = x$$ $$y(ln(x) + 1) = x$$ $$y = \frac{x}{1 + ln(x)}$$

Step 3: Differentiate y with respect to x

Now, differentiate $y$ with respect to $x$ using the quotient rule, which states that if $y = \frac{u}{v}$, then $\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$. Here, $u = x$ and $v = 1 + ln(x)$. So, $\frac{du}{dx} = 1$ and $\frac{dv}{dx} = \frac{1}{x}$. Applying the quotient rule: $$\frac{dy}{dx} = \frac{(1 + ln(x))(1) - x(\frac{1}{x})}{(1 + ln(x))^2}$$ $$\frac{dy}{dx} = \frac{1 + ln(x) - 1}{(1 + ln(x))^2}$$ $$\frac{dy}{dx} = \frac{ln(x)}{(1 + ln(x))^2}$$

Step 4: Final Answer

Thus, we have shown that: $$\frac{dy}{dx} = \frac{ln(x)}{(1 + ln(x))^2}$$

Final Answer: $\frac{log~x}{(1+log~x)^{2}}$

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Pedagogical Audit
Bloom's Analysis: This is an UNDERSTAND question because the student needs to understand the properties of logarithms and differentiation to solve the problem. They must manipulate the given equation and apply the quotient rule correctly.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the student to apply specific procedures like taking logarithms, rearranging equations, and using the quotient rule for differentiation.
Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. The question directly tests the student's knowledge of logarithmic differentiation and the application of differentiation rules, which are core concepts in the syllabus.

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