## amilapsn one year ago Challenge Question: $\color{blueviolet}{\Large\sf{Differentiate\ w.r.t.\ x}}$

1. amilapsn

$\color{blueviolet}{\large\sf{\frac{1}{x^{\alpha-\beta}+x^{\gamma-\beta}+1}+\frac{1}{x^{\gamma-\alpha}+x^{\beta-\alpha}+1}+\frac{1}{x^{\beta-\gamma}+x^{\alpha-\gamma}+1}}}$

2. anonymous

For non-zero $$x$$, the derivative is $$0$$.

3. anonymous

$\frac{1}{x^{\alpha-\beta}+1+x^{\gamma-\beta}}=\frac{x^\beta}{x^\alpha+x^\beta+x^\gamma}$ Do the same with the other two terms, you have $\frac{x^\alpha+x^\beta+x^\gamma}{x^\alpha+x^\beta+x^\gamma}=1$

4. Empty

Even if x=0 you aren't dividing by zero, so I don't think that matters @sithsandgiggles

5. anonymous

But $$\dfrac{0}{0}$$ is an indeterminate form. It certainly does matter if $$x=0$$ in the rewritten expression.

6. Empty

Yeah but the rewritten expression isn't the original one, it's just a thing you made up to solve the problem.

7. anonymous

Without rewriting: Let's just consider these two terms. $\frac{1}{x^{\alpha-\beta}+x^{\gamma-\beta}+1}+\frac{1}{x^{\alpha-\gamma}+x^{\beta-\gamma}+1}$ Suppose $$\alpha<\beta<\gamma$$. Then $$\gamma-\beta>0$$, but $$\beta-\gamma<0$$. So if $$x=0$$, $$x^{\gamma-\beta}=0$$, but $$x^{\beta-\gamma}$$ is undefined. I maintain the original function isn't defined for $$x=0$$, but even if it was that wouldn't imply that the derivative exists at $$x=0$$.