Study23
Hmm.. I'm having difficulties differentiating this function: \(\ \frac{x}{x+\frac{c}{x}} \). Help, please!



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

Study23
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@Mathmuse The fraction is \(\ \Huge \frac{c}{x}.\)

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

Study23
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Could you help me with that?

Study23
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Could I substitute the fraction with \(\ \huge c^{x} ? \)

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

Study23
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So how I go use the quotient rule to differentiate this function?

Jonask
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yes

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

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

Mathmuse
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both will work

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

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

Study23
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Where did \(\ \Huge x^2+c \) come from??

Jonask
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@Mathmuse didnot write x^2 as nominator

Study23
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?

Mathmuse
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for shame

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

Study23
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So how does that become x^2+c?

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

Jonask
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so we derived this