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

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Answer 1 · 2015-09-25T22:25:30

Refer to explanation

First we find the derivative of $(2x)/logx$ hence we have that

$((2x)/logx)'=((2x)'*logx-2x(logx)')/(logx)^2=(2*logx-2)/(logx)^2$

and then the derivative of $x/(logx)^2$ 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

$f'(x)=(2*(logx)^2-3logx+2)/(logx)^3$

Remarks

1).The derivative of a quotient function is

$(f(x)/g(x))=(f'(x)*g(x)-f(x)*g'(x))/(g(x))^2$

2).The derivative of logx is $(logx)'=1/x$

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