Challenge Question:
$$\color{blueviolet}{\Large\sf{Differentiate\ w.r.t.\ x}}$$

Question

Challenge Question:
$$\color{blueviolet}{\Large\sf{Differentiate\ w.r.t.\ x}}$$

amilapsn · Answer

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

anonymous · Answer

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

anonymous · Answer

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

empty · Answer

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

anonymous · Answer

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

empty · Answer

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

anonymous · Answer

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 $x=0$.