Test Index

UPSC NDA 2 2022 Mathematics Solved Paper

Question : 89 of 120

Marks: +1, -0

If y = ln2

\left(\frac{x^2-x+1}{x^2+x+1}\right)

\frac{\text{dy}}{\text{dx}}

at x = 0 equal to ?

Solution:

Concept:

The chain rule and logarithmic differentiation are applied to find the derivative of a composite function.

Explanation:

Given

y = \ln^2 \left( \frac{x^2-x+1}{x^2+x+1} \right)

First, apply the chain rule:

\frac{dy}{dx} = 2 \ln \left( \frac{x^2-x+1}{x^2+x+1} \right) \cdot \frac{d}{dx}\left[ \ln \left( \frac{x^2-x+1}{x^2+x+1} \right) \right]

Then, differentiate the inner logarithm:

\frac{d}{dx} \ln u = \frac{1}{u} \frac{du}{dx}

, where

u = \frac{x^2-x+1}{x^2+x+1}

So,

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

Now differentiate

u

using the quotient rule:

\frac{d}{dx} \left( \frac{x^2-x+1}{x^2+x+1} \right) = \frac{(2x-1)(x^2+x+1) - (x^2-x+1)(2x+1)}{(x^2+x+1)^2}

Substitute

x=0

\ln \left( \frac{1}{1} \right) = \ln 1 = 0

, and the derivative fraction becomes

\frac{(-1)(1) - (1)(1)}{1^2} = -2

Thus,

\frac{dy}{dx} = 2(0) \cdot (1) \cdot (-2) = 0

Answer:

0. So, option B is correct.

Go to Question:

Prev Question

Next Question