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

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amilapsn
 one year ago
Best ResponseYou've already chosen the best response.0\[\color{blueviolet}{\large\sf{\frac{1}{x^{\alpha\beta}+x^{\gamma\beta}+1}+\frac{1}{x^{\gamma\alpha}+x^{\beta\alpha}+1}+\frac{1}{x^{\beta\gamma}+x^{\alpha\gamma}+1}}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For nonzero \(x\), the derivative is \(0\).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{x^{\alpha\beta}+1+x^{\gamma\beta}}=\frac{x^\beta}{x^\alpha+x^\beta+x^\gamma}\] 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
 one year ago
Best ResponseYou've already chosen the best response.0Even if x=0 you aren't dividing by zero, so I don't think that matters @sithsandgiggles

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But \(\dfrac{0}{0}\) is an indeterminate form. It certainly does matter if \(x=0\) in the rewritten expression.

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Yeah but the rewritten expression isn't the original one, it's just a thing you made up to solve the problem.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Without 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 \(x=0\).
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