How do I find the derivative of
ln(ln(1-x))? I know that the derivative of ln(ln(x)) would be 1/xlnx.
Would the answer be 1/(1-x)ln(1-x)?
almost. You forgot a final * (-1) for the derivative of (1-x).
To find the derivative of ln(ln(1-x)), you can use the chain rule.
Let's break it down step by step:
1. Start with the function ln(ln(1-x)).
2. Apply the chain rule:
- The derivative of the outer function ln(u) is 1/u.
- Multiply by the derivative of the inner function ln(1-x).
3. To find the derivative of ln(1-x), we need to apply the chain rule again:
- The derivative of ln(v) is 1/v.
- Multiply by the derivative of the inner function 1-x.
4. The derivative of 1-x is -1, since the derivative of a constant (in this case, x) is always 0.
Putting it all together, we have:
d/dx [ln(ln(1-x))] = (1/(ln(1-x))) * (1/(1-x)) * (-1)
Simplifying this expression gives us:
d/dx [ln(ln(1-x))] = -1/(ln(1-x)(1-x))
So, the derivative of ln(ln(1-x)) is -1/(ln(1-x)(1-x)). Therefore, the answer you provided (1/(1-x)ln(1-x)) is not correct.