Would you dare to differentiate this?

Calculus Level 3

If $y =\dfrac { 1 }{ 1+{ x }^{ b-a }+{ x }^{ c-a } } + \dfrac { 1 }{ 1+{ x }^{ a-b }+{ x }^{ c-b } } + \dfrac { 1 }{ 1+{ x }^{ b-c }+{ x }^{ a-c } }$ , find $\dfrac{dy}{dx}$ .

{ x }^{ a+b+c }

x

2 solutions

Rishabh Jain
Jul 22, 2016

$y= \displaystyle\sum_{\text{cyc}}\dfrac{1}{1+x^{b-a}+x^{c-a}}$

$= \displaystyle\sum_{\text{cyc}}\dfrac{x^a}{x^a+x^{b}+x^{c}}=\dfrac{\displaystyle\sum_{\text{cyc}}x^a }{x^a+x^b+x^c}=1$

$\therefore \dfrac{\mathrm{d}y}{\mathrm dx}= \dfrac{\mathrm{d}(1)}{\mathrm dx}=\boxed 0$

Aman Dubey
Jul 22, 2016

Rearrange it so that we get $y = 1$ (Do it !) now you know.

Would you dare to differentiate this?

2 solutions

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