# How can i find the derivative of f(x)= x÷((x^(1/2) -1) ??

Mar 30, 2018

See process below

#### Explanation:

$f \left(x\right) = \frac{x}{\sqrt{x} - 1}$

Apply cuotient derivative rule

h/g=(h´·g-h·g´)/g^2 where $h = x$ and $g = \sqrt{x} - 1$.

f´(x)=(1·(sqrtx-1)-x·1/(2sqrtx))/(sqrtx-1)^2=(sqrtx-1-1/2sqrtx)/(x-2sqrtx+1)=(1/2sqrtx-1)/(x-2sqrtx+1)

Mar 30, 2018

$\frac{1}{\sqrt{x} - 1} - \frac{\sqrt{x}}{2 {\left(\sqrt{x} - 1\right)}^{2}}$

#### Explanation:

$f \left(x\right) = \frac{x}{\sqrt{x} - 1}$

To find the derivertive of $f \left(x\right)$ we need the quotient-rule

$f \left(x\right) = \frac{k \left(x\right)}{g \left(x\right)}$
$f ' \left(x\right) = \frac{g \left(x\right) k ' \left(x\right) - k \left(x\right) g ' \left(x\right)}{g \left(x\right)} ^ 2$
k(x)=x;" "k'(x)=1
g(x)=sqrt(x)-1=x^(1/2)-1;" "g'(x)=1/2x^(-1/2)=1/(2sqrt(x))

$f ' \left(x\right) = \frac{\left(\sqrt{x} - 1\right) \cdot \left(1\right) - \left(x\right) \cdot \left(\frac{1}{2 \sqrt{x}}\right)}{\sqrt{x} - 1} ^ 2$
$= \frac{\sqrt{x} - 1 - \frac{x}{2 \sqrt{x}}}{\sqrt{x} - 1} ^ 2 = \frac{1}{\sqrt{x} - 1} - \frac{\sqrt{x}}{2 {\left(\sqrt{x} - 1\right)}^{2}}$

I'm sure you will do great:D
If you have any questions, feel free to ask:)