Differentiation Question 143
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
Options:
A) 0
B) 1
C) $ (\alpha +\beta +\gamma ){x^{\alpha +\beta +\gamma -1}} $
D) None of these
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Answer:
Correct Answer: A
Solution:
[a] 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. $
BETA
