Question

If \( y = \frac{x^2}{1 + x^{b-a} + x^{c-a}} + \frac{x^2}{1 + x^{a-b} + x^{c-b}} + \frac{x^2}{1 + x^{a-c} + x^{b-c}} \), then \( \frac{dy}{dx} \) is:

• A. 1
• B. 2x
• C. \( x^a + x^b + x^c \)
• D. \( \frac{1}{x^a + x^b + x^c} \)

Hint:

When faced with symmetric expressions involving exponents, try to homogenize the denominators by multiplying by a suitable factor.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Simplify the expression for \( y \) by multiplying the numerator and denominator of each term by the powers of \( x \) necessary to clear the negative exponents.

Step 2: Detailed Explanation:
Let the terms be \( T_1, T_2, T_3 \).
\( T_1 = \frac{x^2}{1 + \frac{x^b}{x^a} + \frac{x^c}{x^a}} = \frac{x^2 \cdot x^a}{x^a + x^b + x^c} = \frac{x^{a+2}}{x^a + x^b + x^c} \)
\( T_2 = \frac{x^2}{1 + \frac{x^a}{x^b} + \frac{x^c}{x^b}} = \frac{x^2 \cdot x^b}{x^b + x^a + x^c} = \frac{x^{b+2}}{x^a + x^b + x^c} \)
\( T_3 = \frac{x^2}{1 + \frac{x^a}{x^c} + \frac{x^b}{x^c}} = \frac{x^2 \cdot x^c}{x^c + x^a + x^b} = \frac{x^{c+2}}{x^a + x^b + x^c} \)
Summing these up:
\( y = \frac{x^{a+2} + x^{b+2} + x^{c+2}}{x^a + x^b + x^c} = \frac{x^2(x^a + x^b + x^c)}{x^a + x^b + x^c} = x^2 \)

Step 3: Final Answer:
Since \( y = x^2 \), the derivative is \( \frac{dy}{dx} = 2x \). This corresponds to option (B).