If `I_(n)=(d^(n))/(dx^(n))(x^(n)lnx)`, then the value of `(1)/(50)(I_(7)-7I_(6))` is equal to

Answer

To solve the problem, we need to evaluate the expression \( I_n = \frac{d^n}{dx^n}(x^n \ln x) \) and find the value of \( \frac{1}{50}(I_7 - 7I_6) \).

### Step 1: Calculate \( I_n \)

We start with the expression for \( I_n \):
\[ I_n = \frac{d^n}{dx^n}(x^n \ln x) \]

Using the product rule for differentiation, we can express \( I_n \) as:
\[ I_n = \frac{d^n}{dx^n}(x^n) \cdot \ln x + x^n \cdot \frac{d^n}{dx^n}(\ln x) \]

### Step 2: Differentiate \( x^n \)

The \( n \)-th derivative of \( x^n \) is given by:
\[ \frac{d^n}{dx^n}(x^n) = n! \quad \text{(for \( n \geq 0 \))} \]

### Step 3: Differentiate \( \ln x \)

The \( n \)-th derivative of \( \ln x \) can be calculated using the formula:
\[ \frac{d^n}{dx^n}(\ln x) = \frac{(-1)^{n-1}(n-1)!}{x^n} \quad \text{(for \( n \geq 1 \))} \]

### Step 4: Substitute back into \( I_n \)

Substituting these derivatives back into the expression for \( I_n \):
\[ I_n = n! \ln x + x^n \cdot \frac{(-1)^{n-1}(n-1)!}{x^n} \]
\[ I_n = n! \ln x - (-1)^{n-1}(n-1)! \]

### Step 5: Calculate \( I_6 \) and \( I_7 \)

Now, we can find \( I_6 \) and \( I_7 \):
\[ I_6 = 6! \ln x - (-1)^{5} \cdot 5! = 720 \ln x + 120 \]
\[ I_7 = 7! \ln x - (-1)^{6} \cdot 6! = 5040 \ln x - 720 \]

### Step 6: Compute \( I_7 - 7I_6 \)

Now, we compute \( I_7 - 7I_6 \):
\[ I_7 - 7I_6 = (5040 \ln x - 720) - 7(720 \ln x + 120) \]
\[ = 5040 \ln x - 720 - 5040 \ln x - 840 \]
\[ = -720 - 840 = -1560 \]

### Step 7: Calculate \( \frac{1}{50}(I_7 - 7I_6) \)

Finally, we calculate:
\[ \frac{1}{50}(I_7 - 7I_6) = \frac{1}{50}(-1560) = -31.2 \]

### Final Answer

Thus, the value of \( \frac{1}{50}(I_7 - 7I_6) \) is:
\[ \boxed{-31.2} \]