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UPSC NDA 2 2019 Mathematics Solved Paper

Question Numbers: 84-85

Directions: Read the following information and answer the two items that follow:

Consider the equation

x^{y} = e^{x-y}

}

Question : 84 of 120

Marks: +1, -0

What is

\frac{dy}{dx}

at x = 1 equal to?

Solution:

Concept:

Use logarithmic differentiation, properties of logarithms, and standard derivative rules.

Explanation:

Given

x^y = e^{x-y}

, take natural log on both sides:

y \ln x = (x-y) \ln e = x - y

Rearrange to express

x

in terms of

y

x = y(1 + \ln x)

(Equation 1).

Differentiate both sides with respect to

x

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

Solve for

\frac{dy}{dx}

\frac{dy}{dx} = \frac{1 - \frac{y}{x}}{1 + \ln x}

From Equation 1,

\frac{y}{x} = \frac{1}{1 + \ln x}

. Substitute:

\frac{dy}{dx} = \frac{1 - \frac{1}{1 + \ln x}}{1 + \ln x} = \frac{\frac{\ln x}{1 + \ln x}}{1 + \ln x} = \frac{\ln x}{(1 + \ln x)^2}

x = 1

\ln 1 = 0

, so

\frac{dy}{dx} = 0

Answer:

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