If \( y=\displaystyle\frac{x^2}{1+x^{b-a}+x^{c-a}}+\displaystyle\frac{x^2}{1+x^{a-b}+x^{c-b}}+\displaystyle\frac{x^2}{1+x^{a-c}+x^{b-c}} \), then \( \displaystyle\frac{dy}{dx} \) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 2x \)
Multiplying the numerator and denominator of the three fractions by \( x^a \), \( x^b \) and \( x^c \) respectively gives them a common denominator. For the first one:
\( \displaystyle \frac{1}{1+x^{b-a}+x^{c-a}}=\frac{x^a}{x^a+x^b+x^c} \)
Similarly for the other two, so
\( \displaystyle y=x^2\cdot\frac{x^a+x^b+x^c}{x^a+x^b+x^c} \)
\( y=x^2 \)
\( \displaystyle \frac{dy}{dx}=2x \)