Choose the correct alternative in the following:
Question:

Choose the correct alternative in the following:

If, $y=\frac{1}{1+x^{a-b}+x^{c-b}}+\frac{1}{1+x^{b-c}+x^{a-c}} \cdot+\frac{1}{1+x^{b-a}+x^{c-a}}$, then $\frac{d y}{d x}$ is equal to

A. 1

B. $(a+b-c)^{x^{a+b+c-1}}$

C. 0

D. none of these

Solution:

$y=\frac{1}{1+x^{2-b}+x^{c-b}}+\frac{1}{1+x^{b-c}+x^{a-c}}+\frac{1}{1+x^{b-a}+x^{c-a}}$

$\Rightarrow y=\frac{1}{1+\frac{x^{2}}{x^{b}}+\frac{x^{c}}{x^{b}}}+\frac{1}{1+\frac{x^{b}}{x^{c}}+\frac{x^{a}}{x^{c}}}+\frac{1}{1+\frac{x^{b}}{x^{a}}+\frac{x^{c}}{x^{a}}}$

$\Rightarrow y=\frac{x^{b}}{x^{b}+x^{a}+x^{c}}+\frac{x^{c}}{x^{c}+x^{b}+x^{a}}+\frac{x^{a}}{x^{a}+x^{b}+x^{c}}$

$\Rightarrow y=\frac{x^{a}+x^{b}+x^{c}}{x^{a}+x^{b}+x^{c}}=1$

Differentiating w.r.t $x$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=0$

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