Question

$if \: y \: = \: \frac{1}{1 + {x}^{b - a} + {x}^{c - a} } + \frac{1}{1 + {x}^{a - b} + {x}^{c - b} } + \frac{1}{1 + {x}^{a - c} + {x}^{b - c} } \\ \: then \: \frac{dy}{dx} =$
(a)
$0$
(b)
$1$
(c)
$(a + b + c) {x}^{a + b + c - 1}$
(d)
$none \: of \: these$
bahut mehnat ki hai type karne mein correct answer hi dena​

Answer 1

$y = \frac{1}{1 + {x}^{b - a} + {x}^{c - a} } + \frac{1}{1 + {x}^{ a- b} + {x}^{c - b} } + \frac{1}{1 + {x}^{a - c} + {x}^{b - c} } \\ y= \frac{1}{1 + \frac{ {x}^{b} }{ {x}^{a} } + \frac{ {x}^{c} }{ {x}^{a} } } + \frac{1}{1 + \frac{ {x}^{a} }{ {x}^{b} } + \frac{ {x}^{c} }{ {x}^{b} } } + \frac{1}{1 + \frac{ {x}^{a} }{ {x}^{c} } + \frac{ {x}^{b} }{ {x}^{c} } } \\ = \frac{ {x}^{a} }{ {x}^{a} + {x}^{b} + {x}^{c} } + \frac{ {x}^{b} }{ {x}^{b} + {x}^{a} + {x}^{c} } + \frac{ {x}^{c} }{ {x}^{c} + {x}^{a} + {x}^{b} } \\ = \frac{ {x}^{a} + {x}^{b} + {x}^{c} }{ {x}^{a} + {x}^{b} + {x}^{c} } \\ y= 1 \\ \: differentiate \: w.r.t \: \: \: x \\ \frac{dy}{dx} = 0$

Answer 2

Answer:

am 14...

am from Karnataka...wau?