find the derivative of ln(sqrt((1-ln(x))/(1+ln(x)))

It looks messy, but you have

y = ln(sqrt((1-ln(x)) - ln(1+ln(x)))
= 1/2[ ln(1-lnx) - ln(1+lnx) ]
Now, using the chain rule, that is
y' = 1/2 [ 1/(1-lnx)*(-1/x) - 1/(1+lnx)*(1/x) ]
= 1/2 * 1/(x(lnx-1)(lnx+1))

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ln(sqrt((1-ln(x))/(1+ln(x))))
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Find derivative

You will see solution step-by-step.

To find the derivative of ln(sqrt((1-ln(x))/(1+ln(x))), we can use the chain rule.

First, let's simplify the expression to make it easier to differentiate.

ln(sqrt((1-ln(x))/(1+ln(x)))) can be rewritten as 1/2 * ln((1-ln(x))/(1+ln(x))).

Now, let's apply the chain rule:

1. Start by finding the derivative of the outer function, which is ln(u), where u = (1-ln(x))/(1+ln(x)). The derivative of ln(u) is 1/u times the derivative of u.

2. Next, find the derivative of the inner function, u = (1-ln(x))/(1+ln(x)). To differentiate this, we'll use the quotient rule.

a. Let f(x) = 1-ln(x) and g(x) = 1+ln(x).
b. Apply the quotient rule: (f'(x)g(x) - g'(x)f(x)) / g(x)^2.
c. Differentiate f(x) and g(x) step by step:
- Differentiate f(x): f'(x) = -1/x.
- Differentiate g(x): g'(x) = 1/x.
d. Plug these values into the quotient rule:

( (-1/x) * (1 + ln(x)) - (1/x) * (1 - ln(x)) ) / (1 + ln(x))^2

3. Multiply the result obtained in step 2 with the derivative of the outer function:

1/u * ( (-1/x) * (1 + ln(x)) - (1/x) * (1 - ln(x)) ) / (1 + ln(x))^2

4. Simplify the expression:

1/(2u) * ( (-1/x) * (1 + ln(x)) - (1/x) * (1 - ln(x)) ) / (1 + ln(x))^2

Now, combine like terms if necessary and simplify the expression further if possible. This revised expression is the derivative of ln(sqrt((1-ln(x))/(1+ln(x)))).