## Study23 3 years ago Hmm.. I'm having difficulties differentiating this function: $$\ \frac{x}{x+\frac{c}{x}}$$. Help, please!

1. Mathmuse

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

2. Study23

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

3. Mathmuse

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

4. Study23

Could you help me with that?

5. Study23

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

6. Mathmuse

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

7. Study23

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

yes

9. Mathmuse

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

10. Mathmuse

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

11. Mathmuse

both will work

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

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

14. Study23

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

@Mathmuse didnot write x^2 as nominator

16. Study23

?

17. Mathmuse

for shame

$x+\frac{ c }{ x }=\frac{ x^2+c }{ x }$

19. Study23

So how does that become x^2+c?

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