anonymous
  • anonymous
Hmm.. I'm having difficulties differentiating this function: \(\ \frac{x}{x+\frac{c}{x}} \). Help, please!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
function of x. c and z are constants. Direct quotient rule
anonymous
  • anonymous
@Mathmuse The fraction is \(\ \Huge \frac{c}{x}.\)
anonymous
  • anonymous
my bad, could not see it properly. In that case, get a common denominator of x on the bottom.

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anonymous
  • anonymous
Could you help me with that?
anonymous
  • anonymous
Could I substitute the fraction with \(\ \huge c^{-x} ? \)
anonymous
  • anonymous
\[\frac{x}{x + \frac{c}{x}}=\frac{x}{(\frac{x^2+c}{x})}=\frac{x}{1}\frac{x}{(x^2+c)}\]
anonymous
  • anonymous
So how I go use the quotient rule to differentiate this function?
anonymous
  • anonymous
yes
anonymous
  • anonymous
right. Or, because i always mess up the quotient rule, convert it to product rule
anonymous
  • anonymous
\[\frac{x}{x^2+c}=x*(x^2+c)^{-1}\]
anonymous
  • anonymous
both will work
anonymous
  • anonymous
\[(\frac{ x^2 }{ x^2+c })'=\frac{ (x^2)'(x^2+c)-(x^2+c)'x^2 }{ (x^2+c)^2 }\]
anonymous
  • anonymous
\[\frac{ 2x(x^2+c)-2x(x^2) }{ (x^2+c)^2 }=\frac{ 2xc }{ (x^2+c)^2 }\]
anonymous
  • anonymous
Where did \(\ \Huge x^2+c \) come from??
anonymous
  • anonymous
@Mathmuse didnot write x^2 as nominator
anonymous
  • anonymous
?
anonymous
  • anonymous
for shame
anonymous
  • anonymous
\[x+\frac{ c }{ x }=\frac{ x^2+c }{ x }\]
anonymous
  • anonymous
So how does that become x^2+c?
anonymous
  • anonymous
\[x \div \frac{ x^2+c }{ x }=x \times \frac{ x }{ x^2+c }=\frac{ x^2 }{ x^2+c }\]
anonymous
  • anonymous
so we derived this

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