Question

If $y=\frac{1}{1+x^{\beta -\alpha }{x}^{\gamma -\alpha }}+\frac{1}{1+x^{\alpha -\beta }+x^{\gamma -\beta }}+\frac{1}{1+x^{\alpha -\gamma }+x^{\beta -\gamma }}$ then $\frac{dy}{dx}$ is equal to

• A. $0$
• B. $1$
• C. $\left(\alpha +\beta +\gamma \right)x^{\alpha +\beta +\gamma -1}$
• D. None of these

Answer 1

Answer: (C)

We have,
$y=\frac{1}{1+\frac{x^{\beta }}{x^{\alpha }}+\frac{x^{\gamma }}{x^{\alpha }}}+\frac{1}{1+\frac{x^{\alpha }}{x^{\beta }}+\frac{x^{\gamma }}{x^{\beta }}}+\frac{1}{1+\frac{x^{\alpha }}{x^{\gamma }}+\frac{x^{\beta }}{x^{\gamma }}}$
$=\frac{x^{\alpha }}{x^{\alpha }+x^{\beta }+x^{\gamma }}+\frac{x^{\beta }}{x^{\alpha }+x^{\beta }+x^{\gamma }}+\frac{x^{\gamma }}{x^{\alpha }+x^{\beta }+x^{\gamma }}$
$=\frac{x^{\alpha }+x^{\beta }+x^{\gamma }}{x^{\alpha }+x^{\beta }+x^{\gamma }}=1$
$\therefore \frac{dy}{dx}=0$