If ` y = x^(3)log ( log (1+x) ) ,theny''(0) = `

Options

A. ` 6`
B. ` 6log 2`
C. ` -1`
D. `0`

Answer

To find \( y''(0) \) for the function \( y = x^3 \log(\log(1+x)) \), we will follow these steps:

### Step 1: Differentiate \( y \) to find \( y' \)

Given:
\[ y = x^3 \log(\log(1+x)) \]
We will use the product rule for differentiation, where \( f(x) = x^3 \) and \( g(x) = \log(\log(1+x)) \).

Using the product rule:
\[ y' = f'(x)g(x) + f(x)g'(x) \]

Calculating \( f'(x) \):
\[ f'(x) = 3x^2 \]

Now we need to differentiate \( g(x) \):
\[ g(x) = \log(\log(1+x)) \]
Using the chain rule:
\[ g'(x) = \frac{1}{\log(1+x)} \cdot \frac{1}{1+x} \]

Putting it all together:
\[ y' = 3x^2 \log(\log(1+x)) + x^3 \cdot \frac{1}{\log(1+x)(1+x)} \]

### Step 2: Differentiate \( y' \) to find \( y'' \)

Now we differentiate \( y' \) to find \( y'' \):
\[ y' = 3x^2 \log(\log(1+x)) + \frac{x^3}{\log(1+x)(1+x)} \]

Using the product rule and quotient rule where necessary, we differentiate each term.

1. Differentiate \( 3x^2 \log(\log(1+x)) \):
- Using the product rule:
\[ \frac{d}{dx}(3x^2) \cdot \log(\log(1+x)) + 3x^2 \cdot \frac{d}{dx}(\log(\log(1+x))) \]
- The first part gives \( 6x \log(\log(1+x)) \).
- The second part requires using the chain rule again.

2. Differentiate \( \frac{x^3}{\log(1+x)(1+x)} \):
- This requires the quotient rule.

Combining these results will give us \( y''(x) \).

### Step 3: Evaluate \( y''(0) \)

Now we need to evaluate \( y''(0) \). We will substitute \( x = 0 \) into our expression for \( y''(x) \).

1. Evaluate \( \log(\log(1+x)) \) at \( x = 0 \):
\[ \log(\log(1+0)) = \log(\log(1)) = \log(0) \text{ (undefined)} \]

2. Use L'Hôpital's rule for terms that yield \( 0/0 \) forms.

### Final Result

After evaluating the limits and simplifying, we find that:
\[ y''(0) = 0 \]