Let's differentiate the function $y = \frac{\ln(x) - 1}{(\ln(x))^2}$.

We'll use the quotient rule for differentiation, which states that if you have a function of the form $\frac{u(x)}{v(x)}$, then the derivative is given by:

$\frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x) v(x) - u(x) v'(x)}{(v(x))^2}$

Step 1: Define $u(x)$ and $v(x)$

• $u(x) = \ln(x) - 1$
• $v(x) = (\ln(x))^2$

Step 2: Compute the derivatives $u'(x)$ and $v'(x)$

• $u'(x) = \frac{d}{dx} \left( \ln(x) - 1 \right) = \frac{1}{x}$ (since the derivative of $\ln(x)$ is $\frac{1}{x}$ and the derivative of a constant is 0)
• $v'(x) = \frac{d}{dx} \left( (\ln(x))^2 \right) = 2 \ln(x) \cdot \frac{1}{x} = \frac{2 \ln(x)}{x}$ (using the chain rule)

Step 3: Apply the quotient rule

Now we apply the quotient rule:

$y' = \frac{u'(x) v(x) - u(x) v'(x)}{(v(x))^2}$

Substitute $u(x)$, $u'(x)$, $v(x)$, and $v'(x)$ into the formula:

$y' = \frac{\frac{1}{x} \cdot (\ln(x))^2 - (\ln(x) - 1) \cdot \frac{2 \ln(x)}{x}}{(\ln(x))^4}$

Step 4: Simplify the numerator

Simplify the expression:

$y' = \frac{(\ln(x))^2 - 2 \ln(x) (\ln(x) - 1)}{x (\ln(x))^4}$

Now distribute the terms in the numerator:

$y' = \frac{(\ln(x))^2 - 2 (\ln(x))^2 + 2 \ln(x)}{x (\ln(x))^4}$

Combine like terms:

$y' = \frac{- (\ln(x))^2 + 2 \ln(x)}{x (\ln(x))^4}$

Step 5: Final simplified result

Thus, the derivative is:

$y' = \frac{\ln(x) \cdot (2 - \ln(x))}{x (\ln(x))^4}$

This is the derivative of $y = \frac{\ln(x) - 1}{(\ln(x))^2}$.

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Do you want further details or have any questions?

Here are 5 related questions:

1. How do you apply the product rule in differentiation?
2. What are the steps to simplify expressions after using the quotient rule?
3. How does the chain rule apply when differentiating logarithmic functions?
4. Can you explain how to differentiate $\ln(x)^n$?
5. How would you differentiate $\ln(x) - 1$ alone?

Tip: When using the quotient rule, always simplify both the numerator and denominator before finalizing the expression to avoid errors.