JAMIA 2024 — Mathematics PYQ

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Question

JAMIA 2024 — Mathematics PYQ

JAMIA | Mathematics | 2024

If $x^{y} \cdot y^{x} = 16$ , then $\frac{d y}{d x}_{(2, 2)}$ is

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Explanation

Solution

1. Take the Natural Logarithm on both sides:

Given the equation $x^y \cdot y^x = 16$ , take $\ln$ of both sides:

\ln(x^y \cdot y^x) = \ln(16)

Using the properties of logarithms ( $\ln(ab) = \ln a + \ln b$ and $\ln a^b = b \ln a$ ):

y \ln x + x \ln y = \ln(16)

2. Differentiate with respect to $x$ :

We apply the Product Rule $(\frac{d}{dx}[uv] = u\frac{dv}{dx} + v\frac{du}{dx})$ to both terms:

\frac{d}{dx}(y \ln x) + \frac{d}{dx}(x \ln y) = \frac{d}{dx}(\ln 16)

Differentiating each part:

\left( y \cdot \frac{1}{x} + \ln x \cdot \frac{dy}{dx} \right) + \left( x \cdot \frac{1}{y} \cdot \frac{dy}{dx} + \ln y \cdot 1 \right) = 0

3. Isolate $\frac{dy}{dx}$ :

Group the terms containing $\frac{dy}{dx}$ on one side:

\frac{dy}{dx} \ln x + \frac{x}{y} \frac{dy}{dx} = -\frac{y}{x} - \ln y

Factor out $\frac{dy}{dx}$ :

\frac{dy}{dx} \left( \ln x + \frac{x}{y} \right) = -\left( \frac{y}{x} + \ln y \right)

Solve for the derivative:

\frac{dy}{dx} = -\frac{\frac{y}{x} + \ln y}{\ln x + \frac{x}{y}}

4. Evaluate at the point $(2, 2)$ :

Substitute $x = 2$ and $y = 2$ into the expression:

\left.\frac{dy}{dx}\right|_{(2,2)} = -\frac{\frac{2}{2} + \ln 2}{\ln 2 + \frac{2}{2}}

Simplify the fractions:

\left.\frac{dy}{dx}\right|_{(2,2)} = -\frac{1 + \ln 2}{\ln 2 + 1}

Since the numerator and denominator are identical, they cancel out:

\left.\frac{dy}{dx}\right|_{(2,2)} = -1

Final Answer

\left.\frac{dy}{dx}\right|_{(2,2)} = -1

Choose the correct answer:

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