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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 ______________ .
Options
1
\[\left( a + b + c \right)^{x^{a + b + c - 1}}\]
0
none of these
Solution
`0`
\[\text { We have }, \]
\[y = \frac{1}{1 + x^{a - b} + x^{c - b}} + \frac{1}{1 + x^{b - c} + x^{a - c}} + \frac{1}{1 + x^{b - a} + x^{c - a}}\]
\[ = \frac{1}{1 + \frac{x^a}{x^b} + \frac{x^c}{x^b}} + \frac{1}{1 + \frac{x^b}{x^c} + \frac{x^a}{x^c}} + \frac{1}{1 + \frac{x^b}{x^a} + \frac{x^c}{x^a}}\]
\[ = \frac{x^b}{x^b + x^a + x^c} + \frac{x^c}{x^c + x^b + x^a} + \frac{x^a}{x^a + x^b + x^c}\]
\[ = \frac{x^b}{x^a + x^b + x^c} + \frac{x^c}{x^a + x^b + x^c} + \frac{x^a}{x^a + x^b + x^c}\]
\[ = \frac{x^b + x^c + x^a}{x^a + x^b + x^c}\]
\[ = \frac{x^a + x^b + x^c}{x^a + x^b + x^c}\]
\[ = 1\]
\[ \therefore \frac{dy}{dx} = \frac{d}{dx}\left( 1 \right) = 0\]
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