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function of x. c and z are constants.
Direct quotient rule

my bad, could not see it properly. In that case, get a common denominator of x on the bottom.

Could you help me with that?

Could I substitute the fraction with \(\ \huge c^{-x} ? \)

\[\frac{x}{x + \frac{c}{x}}=\frac{x}{(\frac{x^2+c}{x})}=\frac{x}{1}\frac{x}{(x^2+c)}\]

So how I go use the quotient rule to differentiate this function?

yes

right. Or, because i always mess up the quotient rule, convert it to product rule

\[\frac{x}{x^2+c}=x*(x^2+c)^{-1}\]

both will work

\[(\frac{ x^2 }{ x^2+c })'=\frac{ (x^2)'(x^2+c)-(x^2+c)'x^2 }{ (x^2+c)^2 }\]

\[\frac{ 2x(x^2+c)-2x(x^2) }{ (x^2+c)^2 }=\frac{ 2xc }{ (x^2+c)^2 }\]

Where did \(\ \Huge x^2+c \) come from??

for shame

\[x+\frac{ c }{ x }=\frac{ x^2+c }{ x }\]

So how does that become x^2+c?

\[x \div \frac{ x^2+c }{ x }=x \times \frac{ x }{ x^2+c }=\frac{ x^2 }{ x^2+c }\]

so we derived this