Question

If $y=\frac{1}{1+x^{a-b}{+}^{c-b}}+\frac{1}{1+x^{b-c}+x^{a-c}}+\frac{1}{1+x^{b-a}+x^{c-a}}$, then $\frac{dy}{dx}$ is equal to
(a) 1
(b) ${\left(a+b+c\right)}^{x^{a+b+c-1}}$
(c) 0
(d) none of these

Answer 1

(c) 0


$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ y=\frac{1}{1+x^{a-b}+x^{c-b}}+\frac{1}{1+x^{b-c}+x^{a-c}}+\frac{1}{1+x^{b-a}+x^{c-a}} \\ =\frac{1}{1+{\displaystyle \frac{x^a}{x^b}}+{\displaystyle \frac{x^c}{x^b}}}+\frac{1}{1+{\displaystyle \frac{x^b}{x^c}}+{\displaystyle \frac{x^a}{x^c}}}+\frac{1}{1+{\displaystyle \frac{x^b}{x^a}}+{\displaystyle \frac{x^c}{x^a}}} \\ =\frac{x^b}{x^b+x^a+x^c}+\frac{x^c}{x^c+x^b+x^a}+\frac{x^a}{x^a+x^b+x^c} \\ =\frac{x^b}{x^a+x^b+x^c}+\frac{x^c}{x^a+x^b+x^c}+\frac{x^a}{x^a+x^b+x^c} \\ =\frac{x^b+x^c+x^a}{x^a+x^b+x^c} \\ =\frac{x^a+x^b+x^c}{x^a+x^b+x^c} \\ =1 \\ \therefore \frac{dy}{dx}=\frac{d}{dx}\left(1\right)=0 \end{gathered}$$