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