Question

In Exercise, find the derivative of the function.
y = In 5x/x + 2

Answer 1

`y'=(1+(2)/(x)-ln5x )/(x^2+4x+4)`

Explanation:

The function

`y=(ln5x)/(x+2)`

its derivative can be found with the derivative rule of the quotient

if
`y=(a)/(b)`

the derivative is

`y'=(a'b-ab')/(b^2)`

in this case

`a=ln5x`

and

`b=x+2`

thus Using the logarithm derivation:
`(lnu)'=(u')/(u)`, so

`a'=(5)/(5x) =(1)/(x)`

and

`b'=1`

the derivative of
`y=(ln5x)/(x+2)` is:

`y'=(a'b-ab')/(b^2)=(((1)/(x) )(x+2)-(ln5x)(1))/((x+2)^2)\\y'=(1+(2)/(x)-ln5x )/(x^2+4x+4)`

Answer 2

The derivative of the function is
`y'=(1+(2)/(x)-ln5x)/(x^2+4x+4)`

Explanation:

Be

`y=(ln5x)/(x+2)`

then, to derivate we have to apply the rule for division and the chain rule, and the division rule:

`y'=(d)/(dx)((ln5x)/(x+2))=((d)/(dx)(ln5x)(x+2)-(ln5x)(d)/(dx)(x+2))/((x+2)^2)=(((1)/(x))(x+2)-ln5x(1))/(x^2+4x+4)=(1+(2)/(x)-ln5x)/(x^2+4x+4)`

which is our derivative, and because, by chain rule we calculated

`(d)/(dx)(ln5x)=(1)/(5x)*5=(1)/(x)`