Question

How do I find the derivative of `ln[x(x^2+1)^2/(2x^3-1)^(1/2)]`?

Answer 1

Use the properties of logarithms and the `d/dxln(x)` rule to get `1/x+(4x)/(x^2+1)-(3x^2)/(2x^3-1)`.

Explanation:

Begin by using the properties of logs to write:
`ln[(x(x^2+1)^2)/(2x^3-1)^(1/2)]=ln(x(x^2+1)^2)-ln(2x^3-1)^(1/2)=ln(x)+ln((x^2+1)^2)-1/2ln(2x^3-1)=ln(x)+2ln(x^2+1)-1/2ln(2x^3-1)`

Now we can take the derivative of this term by term.

Term 1: `ln(x)`
This is the easiest, as the derivative of `ln(x)=1/x`.

Term 2: `2ln(x^2+1)`
We need to use the chain rule for this one. We first apply the derivative of `ln(x)` rule to get `2*1/(x^2+1)`, and then we apply the chain rule to get `2*(2x)/(x^2+1)=(4x)/(x^2+1)`.

Term 3: `1/2ln(2x^3-1)`
We use the same process as term 2. Use `ln(x)` rule to get `1/2*1/(2x^3-1)`, then use the chain rule to get `1/2*(6x^2)/(2x^3-1)=(3x^2)/(2x^3-1)`.

Put all of these together to get the final result:
`d/dxln[(x(x^2+1)^2)/(2x^3-1)^(1/2)]=1/x+(4x)/(x^2+1)-(3x^2)/(2x^3-1)`.