Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

Could you help me with that?

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

yes

Answer 9

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

Answer 10

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

Answer 11

both will work

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

@Mathmuse didnot write x^2 as nominator

Answer 16

?

Answer 17

for shame

Answer 18

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

Answer 19

So how does that become x^2+c?

Answer 20

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

Answer 21

so we derived this