Question

How do you differentiate `y =-ln [ 3+ (9+x^2) / x]`?

Answer 1

`dy/dx=-1/(3+9/x+x)*(1-9/x^2)`

Explanation:

`y=-ln(3+(9+x^2)/x)=-ln(3+9/x+x)`

Use the chain rule `dy/dx=dy/(du)*(du)/dx`

Let `y=-ln(u)` then `dy/(du)=-1/u`
And `u=3+9/x+x` then `(du)/(dx)=-9/x^2+1`

`dy/dx=-1/u*(1-9/x^2)`

Substitute `u=3+9/x+x`

`dy/dx=-1/(3+9/x+x)*(1-9/x^2)`