amilapsn
  • amilapsn
Challenge Question: \[\color{blueviolet}{\Large\sf{Differentiate\ w.r.t.\ x}}\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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amilapsn
  • amilapsn
\[\color{blueviolet}{\large\sf{\frac{1}{x^{\alpha-\beta}+x^{\gamma-\beta}+1}+\frac{1}{x^{\gamma-\alpha}+x^{\beta-\alpha}+1}+\frac{1}{x^{\beta-\gamma}+x^{\alpha-\gamma}+1}}}\]
anonymous
  • anonymous
For non-zero \(x\), the derivative is \(0\).
anonymous
  • 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 \[\frac{x^\alpha+x^\beta+x^\gamma}{x^\alpha+x^\beta+x^\gamma}=1\]

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Empty
  • Empty
Even if x=0 you aren't dividing by zero, so I don't think that matters @sithsandgiggles
anonymous
  • anonymous
But \(\dfrac{0}{0}\) is an indeterminate form. It certainly does matter if \(x=0\) in the rewritten expression.
Empty
  • Empty
Yeah but the rewritten expression isn't the original one, it's just a thing you made up to solve the problem.
anonymous
  • 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 \(x=0\).

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