Illustrate the process for finding the derivative of the function h(x) = ln(x)/x^2.
user
Sure, I'd be happy to help you find the derivative of h(x) = ln(x)/x^2.

1. **Recall the chain rule:** If f(u) and g(x) are both functions, and if u = g(x), then f'(x) = f'(u)g'(x).

2. **Apply the chain rule to h(x).** Here's how:

   - In this case, u = x, f(u) = ln(u), and g(x) = x^2
   
   - So, we'll use the chain rule: h'(x) = [ln(x)]' * (x^2)' = [1/u * u'] * [2x]
     
   - Simplify the first factor: 1/u * u' = (1/x) * 1 = 1/x

   - Take the derivative of the second factor: u' = (1) * (2x) = 2x

   - Therefore, h'(x) = [1/x] * [2x] = 2/x

3. **Notice:** Since we're dealing with ln(x), this is an indeterminate form. In such cases, you can use l'Hopital's rule, which states that if f(x) and g(x) approach infinity or zero together (indeterminate), then f'(x)/g'(x) is equal to [f'(t)/g'(t)] as t → x.

   - Applying this, we can say:
       
     - f'(x) = lim [2/x] = [2/(2x)] = [1/x] = f'(x)/1 ≈ lim [ln(x+t)/(2x)]/[1/(x)]

   - Now, take the common denominator: [ln(x+t)] * x = t * ln(x+t)

   - Simplify: x * ln(x+t) /1 = t * ln(x+t).

   - Thus, 

      lim Ln(x) / x = lim [t * ln(x + t)] / [2x]

      = [t * ds / dx] / [2x] = [t(dx/dx)] / [2x]

   - Substitute t = 1, dt = dx/dx

   - Finally, substitute ln(x+1) = natural logarithm of (x + 1)

     - Thus, the limit is inv(2x) * ln(x+1), where "inv" means the inverse function.

     - So, lim [inv(2x)] = (1/2) * ln(2x).

   - Therefore, lim [ln(x)]' / (2x) = (1/2) * ln(2x) / (2x) = (1/4) * ln(2x).

4. **The final step:** Here, we differentiate the final result again using the product to power rule or u-substitution.

   - However, since we completed the limit calculation in step 3, we can remove the "lim" symbol from these lines.

   - Simply finishing off the differential command with h' is:

       h'(x) = (1/4) * [ln(2x)]' / (2x) = (1/4) * (1/2x) * [2 * ln(2)] + (1/4) * (1/2x * 2x + 1/2 * 2) = 1/2x.

   - Proving that this is, in fact, h'(x).

   - Therefore, h'(x) = 1/2x.
assistant
Illustrate the process for finding the derivative of the function h(x) = ln(x)/x^2.
user
To find the derivative of h(x) = ln(x)/x^2, we can use the quotient rule, which states that for a function f(x)/g(x), its derivative is given by:

(f'(x)g(x) - f(x)g'(x)) / (g(x))^2

In our case, f(x) = ln(x) and g(x) = x^2. So first, let's find their derivatives:

f'(x) = d/dx [ln(x)] = 1/x
g'(x) = d/dx [x^2] = 2x

Now, apply the quotient rule:

h'(x) = ( (1/x)(x^2) - (ln(x))(2x) ) / (x^2)^2

Simplify the expression:

h'(x) = ( x - 2x*ln(x) ) / (x^4)

Factor out an x from the numerator:

h'(x) = (x * (1 - 2*ln(x))) / (x^4)

Cancel out a common factor of x between the numerator and denominator:

h'(x) = (1 - 2*ln(x)) / (x^3)

So the derivative of h(x) = ln(x)/x^2 is h'(x) = (1 - 2*ln(x)) / (x^3).
assistant