A problem on finding dy/dx

Question

If $a+b+c=0$ and $$y=\frac{1}{x^b+x^{-c}+1}+\frac{1}{x^c+x^{-a}+1}+\frac{1}{x^a+x^{-b}+1}$$then $\frac{dy}{dx}$=?

The only way which I can think of solving this is by differentiating each term. However, is there a simpler way?

Answer 1

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

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

Answer 2

Try to simplify $y$ before differentiating, you should get $y=1$

Answer 3

Hint! make $u=x^a$ and $v=x^b$, we have that $a+b+c=0$ so $x^c=?$, replace and simplify.