Question

How do you differentiate `y=ln[x/(2x+3)]^(1/2)`?

Answer 1

Differentiate: `y=ln[x/(2x+3)]^(1/2)`

Use the property `ln(a^c) = (c)ln(a)`

`y=1/2ln(x/(2x+3))`

Use the property `ln(a/b) = ln(a)-ln(b)`

`y=1/2ln(x)-1/2ln(2x+3)`

Now, we may differentiate each term:

`dy/dx=1/(2x)-1/2 2/(2x+3)`

`dy/dx=1/(2x)-1/(2x+3)`

Answer 2

`dy/dx=3/{2x(x+3)}.`

Explanation:

We have, `y=ln{x/(2x+3)}^(1/2).`

Using the familiar rules of Log. Fun., we get,

`y=1/2ln(x/(2x+3))=1/2{lnx-ln(2x+3)}.`

`:. dy/dx=d/dx[1/2{lnx-ln(2x+3)}].`

`=1/2d/dx{lnx-ln(2x+3)},`

`=1/2[d/dx{lnx}-d/dx{ln(2x+3)}],`

`=1/2[1/x-1/(2x+3)d/dx{(2x+3)}],`

`=1/2[1/x-2/(2x+3)],`

`=1/2[{(2x+3)-2x}/{x(2x+3)}],`

`rArr dy/dx=3/{2x(2x+3)}.`