Method of DifferentiationmediumFree

Method of Differentiation — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question

y=\dfrac{1}{1+x^{n-m}+x^{p-m}}+\dfrac{1}{1+x^{m-n}+x^{p-n}}+\dfrac{1}{1+x^{m-p}+x^{n-p}}

, then

\dfrac{dy}{dx}

x=e^{mnp}

is equal to

e^{mnp}

e^{mn/p}

e^{mp/n}

DNone of thesecorrect

Solution

Step 1: Multiply the numerator and denominator of each fraction by

x^{m}

x^{n}

, and

x^{p}

respectively:

\dfrac{1}{1+x^{n-m}+x^{p-m}}=\dfrac{x^{m}}{x^{m}+x^{n}+x^{p}}

\dfrac{1}{1+x^{m-n}+x^{p-n}}=\dfrac{x^{n}}{x^{n}+x^{m}+x^{p}}

\dfrac{1}{1+x^{m-p}+x^{n-p}}=\dfrac{x^{p}}{x^{p}+x^{m}+x^{n}}

Step 2: Since all three fractions share the denominator

x^{m}+x^{n}+x^{p}

, their sum is:

y=\dfrac{x^{m}+x^{n}+x^{p}}{x^{m}+x^{n}+x^{p}}=1

Step 3: Since

y=1

(a constant),

\dfrac{dy}{dx}=0

for all

x

. Therefore:

\left(\dfrac{dy}{dx}\right)_{x=e^{mnp}}=0

Correct answer: (4) None of these

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