Q1382: If $x=e^{\frac{x}{y}}$, then prove that $\frac{dy}{dx}=\frac...

Class CBSE Class 12 Mathematics Derivatives Q #1382

KNOWLEDGE BASED

UNDERSTAND

2 Marks 2025 AISSCE(Board Exam) VSA

If $x=e^{\frac{x}{y}}$, then prove that $\frac{dy}{dx}=\frac{x-y}{x\log x}$.

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Detailed Solution

Step 1: Take the natural logarithm of both sides

Given the equation $x = e^{\frac{x}{y}}$, we take the natural logarithm (ln) of both sides to simplify the exponential term:

Step 2: Simplify the equation using logarithm properties

Using the property $\ln(e^a) = a$, we get: $\ln(x) = \ln(e^{\frac{x}{y}})$ $\ln(x) = \frac{x}{y}$

Step 3: Solve for y in terms of x

Rearrange the equation to solve for $y$: $y = \frac{x}{\ln(x)}$

Step 4: 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 $v = \ln(x)$. $\frac{du}{dx} = 1$ $\frac{dv}{dx} = \frac{1}{x}$ Applying the quotient rule: $\frac{dy}{dx} = \frac{\ln(x) \cdot 1 - x \cdot \frac{1}{x}}{(\ln(x))^2}$ $\frac{dy}{dx} = \frac{\ln(x) - 1}{(\ln(x))^2}$

Step 5: Rewrite the expression in terms of x and y

Since $\ln(x) = \frac{x}{y}$, substitute this back into the expression for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{\frac{x}{y} - 1}{(\frac{x}{y})^2}$ $\frac{dy}{dx} = \frac{\frac{x-y}{y}}{\frac{x^2}{y^2}}$ $\frac{dy}{dx} = \frac{x-y}{y} \cdot \frac{y^2}{x^2}$ $\frac{dy}{dx} = \frac{y(x-y)}{x^2}$

Step 6: Substitute y again to get the final form

Substitute $y = \frac{x}{\ln(x)}$ back into the equation: $\frac{dy}{dx} = \frac{\frac{x}{\ln(x)}(x- \frac{x}{\ln(x)})}{x^2}$ $\frac{dy}{dx} = \frac{x(x- \frac{x}{\ln(x)})}{x^2 \ln(x)}$ $\frac{dy}{dx} = \frac{x(1- \frac{1}{\ln(x)})}{x \ln(x)}$ $\frac{dy}{dx} = \frac{1- \frac{1}{\ln(x)}}{\ln(x)}$ $\frac{dy}{dx} = \frac{\ln(x) - 1}{(\ln(x))^2}$ $\frac{dy}{dx} = \frac{\frac{x}{y} - 1}{(\frac{x}{y})^2} = \frac{y^2}{x^2} \cdot \frac{x-y}{y} = \frac{y(x-y)}{x^2} = \frac{\frac{x}{\ln x}(x- \frac{x}{\ln x})}{x^2} = \frac{x(x- \frac{x}{\ln x})}{x^2 \ln x} = \frac{x- \frac{x}{\ln x}}{x \ln x} = \frac{x(1- \frac{1}{\ln x})}{x \ln x} = \frac{1- \frac{1}{\ln x}}{\ln x} = \frac{\ln x - 1}{(\ln x)^2}$ Now, we need to show that $\frac{\ln x - 1}{(\ln x)^2} = \frac{x-y}{x \ln x}$. $\frac{x-y}{x \ln x} = \frac{x - \frac{x}{\ln x}}{x \ln x} = \frac{x(1 - \frac{1}{\ln x})}{x \ln x} = \frac{1 - \frac{1}{\ln x}}{\ln x} = \frac{\ln x - 1}{(\ln x)^2}$ Thus, $\frac{dy}{dx} = \frac{x-y}{x \ln x}$.

Final Answer: $\frac{dy}{dx}=\frac{x-y}{x\log x}$

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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 such as taking logarithms, using the quotient rule for differentiation, and algebraic manipulation to arrive at the solution.<\/span>

Syllabus Audit: In the context of CBSE Class 12, this is classified as APPLICATION. The question involves implicit differentiation and requires the student to apply the concepts of derivatives and logarithmic functions, which are part of the CBSE Class 12 syllabus.