If \( y=\left[\displaystyle\frac{2}{1+x^{b-a}+x^{c-a}}+\frac{2}{1+x^{c-b}+x^{a-b}}+\frac{2}{1+x^{a-c}+x^{b-c}}\right]^x \), then \( \displaystyle\frac{dy}{dx} \) is equal to :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 2^x\log_e 2 \)
The bracket simplifies to a constant. Multiplying the numerator and denominator of the first term by \( x^a \),
\( \displaystyle \frac{2}{1+x^{b-a}+x^{c-a}}=\frac{2x^a}{x^a+x^b+x^c} \)
In the same way the second term is \( \frac{2x^b}{x^a+x^b+x^c} \) (multiply by \( x^b \)) and the third is \( \frac{2x^c}{x^a+x^b+x^c} \) (multiply by \( x^c \)). Adding:
\( \displaystyle \frac{2(x^a+x^b+x^c)}{x^a+x^b+x^c}=2 \)
So \( y=2^x \), and by the derivative of \( a^x \),
\( \displaystyle \frac{dy}{dx}=2^x\log_e 2 \)