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UPSC NDA 2 2021 Mathematics Solved Paper

Question : 71 of 120

Marks: +1, -0

y = (1 + x)(1 + x^{2})(1 + x^{4})(1 + x^{8})(1 + x^{16})

\frac{dy}{dx}

x = 0

equal to?

Solution:

Concept:

Logarithmic differentiation converts a product into a sum of simpler derivatives.

Explanation:

Let

y = (1+x)(1+x^2)(1+x^4)(1+x^8)(1+x^{16})

x=0

y(0) = 1

Take the natural logarithm of both sides:

\ln y = \ln(1+x) + \ln(1+x^2) + \ln(1+x^4) + \ln(1+x^8) + \ln(1+x^{16})

Differentiate with respect to

x

\frac{1}{y}\frac{dy}{dx} = \frac{1}{1+x} + \frac{2x}{1+x^2} + \frac{4x^3}{1+x^4} + \frac{8x^7}{1+x^8} + \frac{16x^{15}}{1+x^{16}}

Multiply both sides by

y

\frac{dy}{dx} = y\left(\frac{1}{1+x} + \frac{2x}{1+x^2} + \frac{4x^3}{1+x^4} + \frac{8x^7}{1+x^8} + \frac{16x^{15}}{1+x^{16}}\right)

x=0

y(0)=1

All terms except

\frac{1}{1+0}

become zero because each numerator contains a factor of

x

Thus,

\left(\frac{dy}{dx}\right)_{x=0} = 1 \times 1 = 1

Answer:

1

(Option B)

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