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Study23 Group Title

Hmm.. I'm having difficulties differentiating this function: \(\ \frac{x}{x+\frac{c}{x}} \). Help, please!

  • one year ago
  • one year ago

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  1. Mathmuse Group Title
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    function of x. c and z are constants. Direct quotient rule

    • one year ago
  2. Study23 Group Title
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    @Mathmuse The fraction is \(\ \Huge \frac{c}{x}.\)

    • one year ago
  3. Mathmuse Group Title
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    my bad, could not see it properly. In that case, get a common denominator of x on the bottom.

    • one year ago
  4. Study23 Group Title
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    Could you help me with that?

    • one year ago
  5. Study23 Group Title
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    Could I substitute the fraction with \(\ \huge c^{-x} ? \)

    • one year ago
  6. Mathmuse Group Title
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    \[\frac{x}{x + \frac{c}{x}}=\frac{x}{(\frac{x^2+c}{x})}=\frac{x}{1}\frac{x}{(x^2+c)}\]

    • one year ago
  7. Study23 Group Title
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    So how I go use the quotient rule to differentiate this function?

    • one year ago
  8. Jonask Group Title
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    yes

    • one year ago
  9. Mathmuse Group Title
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    right. Or, because i always mess up the quotient rule, convert it to product rule

    • one year ago
  10. Mathmuse Group Title
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    \[\frac{x}{x^2+c}=x*(x^2+c)^{-1}\]

    • one year ago
  11. Mathmuse Group Title
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    both will work

    • one year ago
  12. Jonask Group Title
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    \[(\frac{ x^2 }{ x^2+c })'=\frac{ (x^2)'(x^2+c)-(x^2+c)'x^2 }{ (x^2+c)^2 }\]

    • one year ago
  13. Jonask Group Title
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    \[\frac{ 2x(x^2+c)-2x(x^2) }{ (x^2+c)^2 }=\frac{ 2xc }{ (x^2+c)^2 }\]

    • one year ago
  14. Study23 Group Title
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    Where did \(\ \Huge x^2+c \) come from??

    • one year ago
  15. Jonask Group Title
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    @Mathmuse didnot write x^2 as nominator

    • one year ago
  16. Study23 Group Title
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    ?

    • one year ago
  17. Mathmuse Group Title
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    for shame

    • one year ago
  18. Jonask Group Title
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    \[x+\frac{ c }{ x }=\frac{ x^2+c }{ x }\]

    • one year ago
  19. Study23 Group Title
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    So how does that become x^2+c?

    • one year ago
  20. Jonask Group Title
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    \[x \div \frac{ x^2+c }{ x }=x \times \frac{ x }{ x^2+c }=\frac{ x^2 }{ x^2+c }\]

    • one year ago
  21. Jonask Group Title
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    so we derived this

    • one year ago
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