# How do you differentiate  f(x) = [ (2x)/(log(x)) ] - [ x/(log²(x)) ] ?

Refer to explanation

#### Explanation:

First we find the derivative of $\frac{2 x}{\log} x$ hence we have that

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

and then the derivative of $\frac{x}{\log x} ^ 2$ hence we have that

(x/(logx)^2)'=((x)'*(logx)^2-x((logx)^2)')/(logx)^4= ((logx)^2-2xlogx*(1/x))/(logx)^4=((logx-2))/(logx)^3

Finally we get

$f ' \left(x\right) = \frac{2 \cdot {\left(\log x\right)}^{2} - 3 \log x + 2}{\log x} ^ 3$

Remarks

1).The derivative of a quotient function is

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

2).The derivative of logx is $\left(\log x\right) ' = \frac{1}{x}$