`f(x)=x^2e^xlogx`
differentiate f(x) with respect to x at x = 1

Options

A. 0
B. 1
C. e
D. none of these

Explanation

To differentiate the function \( f(x) = x^2 e^x \log x \) with respect to \( x \), we will use the product rule of differentiation. The product rule states that if you have a product of three functions \( u, v, w \), then the derivative is given by:

\[ f'(x) = u'vw + uv'w + uvw' \]

In our case:
- Let \( u = x^2 \)
- Let \( v = e^x \)
- Let \( w = \log x \)

Now, we will differentiate each of these functions:

1. **Differentiate \( u \)**:
\[ u' = \frac{d}{dx}(x^2) = 2x \]

2. **Differentiate \( v \)**:
\[ v' = \frac{d}{dx}(e^x) = e^x \]

3. **Differentiate \( w \)**:
\[ w' = \frac{d}{dx}(\log x) = \frac{1}{x} \]

Now, we can apply the product rule:

\[ f'(x) = u'vw + uv'w + uvw' \]

Substituting the derivatives and the original functions into the equation:

\[ f'(x) = (2x)(e^x)(\log x) + (x^2)(e^x)(\log x) + (x^2)(e^x)\left(\frac{1}{x}\right) \]

Now, let's simplify the expression:

1. The first term is \( 2x e^x \log x \).
2. The second term is \( x^2 e^x \log x \).
3. The third term simplifies to \( x e^x \).

Combining these terms, we can factor out \( e^x \):

\[ f'(x) = e^x \left( 2x \log x + x^2 \log x + x \right) \]

Now, we can evaluate \( f'(x) \) at \( x = 1 \):

1. Calculate \( \log(1) = 0 \).
2. Substitute \( x = 1 \) into the derivative:

\[ f'(1) = e^1 \left( 2(1)(0) + (1^2)(0) + 1 \right) \]

This simplifies to:

\[ f'(1) = e \left( 0 + 0 + 1 \right) = e \]

Thus, the derivative of \( f(x) \) at \( x = 1 \) is:

\[ \boxed{e} \]