What is the derivative of ln((4x+9)/(3x-5))^1/2?

Question

What is the derivative of ln((4x+9)/(3x-5))^1/2?

Answer 1

Let y = ln((4x+9)/(3x-5))^1/2

First we'll use the rules of log to simplify the problem.

Note that ln a^b = b*ln a.

So it becomes y = (1/2)*ln((4x+9)/(3x-5))

Now use the fact that ln(a/b) = ln a - ln b and it becomes:

y = (1/2)* [ ln(4x+9) - ln(3x-5)]

Now we can take dy/dx:

dy/dx = (1/2)* {  [1/(4x+9)]* 4 - [1/(3x-5)]*3 }

Note we had to use the chain rule for each ln term

Specifically, if  w= ln(4x+9) then to get dw/dx we say let u=4x+9

Then dw/dx = (dw/du)*(du/dx) = [1/u]*(4)

The other term is done similarly.

Let me know if you need more details.

The key thing is that whenever you have logs, look to use the properties of logs to simplify the problem.