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 ______________ .

पर्याय

• 1

• \[\left( a + b + c \right)^{x^{a + b + c - 1}}\]

• 0

• none of these

Answer 1

`0`

 

\[\text { We have }, \]
\[y = \frac{1}{1 + x^{a - b} + x^{c - b}} + \frac{1}{1 + x^{b - c} + x^{a - c}} + \frac{1}{1 + x^{b - a} + x^{c - a}}\]
\[ = \frac{1}{1 + \frac{x^a}{x^b} + \frac{x^c}{x^b}} + \frac{1}{1 + \frac{x^b}{x^c} + \frac{x^a}{x^c}} + \frac{1}{1 + \frac{x^b}{x^a} + \frac{x^c}{x^a}}\]
\[ = \frac{x^b}{x^b + x^a + x^c} + \frac{x^c}{x^c + x^b + x^a} + \frac{x^a}{x^a + x^b + x^c}\]
\[ = \frac{x^b}{x^a + x^b + x^c} + \frac{x^c}{x^a + x^b + x^c} + \frac{x^a}{x^a + x^b + x^c}\]
\[ = \frac{x^b + x^c + x^a}{x^a + x^b + x^c}\]
\[ = \frac{x^a + x^b + x^c}{x^a + x^b + x^c}\]
\[ = 1\]
\[ \therefore \frac{dy}{dx} = \frac{d}{dx}\left( 1 \right) = 0\]