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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

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Solution

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

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