What is the derivative of $ln(2x)/x$ ?

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Answer 1 · 2016-10-29T19:41:57

Let's start by finding the derivative of $ln(2x)$ .

Let $y = lnu$ and $u = 2x$ .

Then $y' = 1/u$ and $u' = 2$ .

$dy/dx = 1/u xx 2 = 2/(2x) = 1/x$

We can now use the quotient rule to differentiate the entire function.

$dy/dx = (1/x xx x - ln(2x) xx 1)/(x)^2$

$dy/dx = (1-ln(2x))/x^2$

Hopefully this helps!

Answer 2 · 2016-10-29T19:50:31

The derivative is $=(1-ln(2x))/x^2$

This is the derivative of a quotient
$(u/v)'=(u'v-uv')/v^2$
so here
$u=ln(2x)$ $=>$ $u'=2/(2x)=1/x$

$v=x$ $=>$ $v'=1$

So, $(ln(2x)/x)'=(x*1/x-1*ln(2x))/x^2$

$=(1-ln(2x))/x^2$

What is the derivative of ln(2x)/xln(2x)/x?