# Can anyone help find the derivatives of a given function? 1) y=(lnx)/(3x-1+x^2) Thank's

### 2 Answers | Add Yours

If you have more than one question, you need to make separate posts.

To find the derivative of the function, we need to use the quotient rule. This means that the derivative is:

`y'={(1/x)(3x-1+x^2)-lnx(3+2x)}/(3x-1+x^2)^2` simplify numerator

`={3x-1+x^2-x(3+2x)lnx}/{x(3x-1+x^2)^2}`

**The derivative is `{3x-1+x^2-x(3+2x)lnx}/{x(3x-1+x^2)^2}` .**

Using the quotient rule,

we have :

[(1/x)(3x-1+x^2) - (lnx)(3+2x)] /(3x-1+x^2)^2

And then we clean this up:

[3-1/x + x - 3lnx -2xlnx]/(3x-1+x^2)^2

You could open the denominator but I think it's best if we leave it at this step.

Again, you could try further but it doesn't seem it would help.