Which of the following could be the sketch graph of `y = (d(xlnx))/dx`

Options

A.
B.
C.
D.

Answer

To solve the problem of finding the sketch graph of \( y = \frac{d}{dx}(x \ln x) \), we will follow these steps:

### Step 1: Differentiate the function
We start with the function \( y = x \ln x \). To find \( \frac{dy}{dx} \), we will use the product rule of differentiation.

The product rule states that if you have two functions \( u \) and \( v \), then:
\[ \frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx} \]

Here, let:
- \( u = x \) (which implies \( \frac{du}{dx} = 1 \))
- \( v = \ln x \) (which implies \( \frac{dv}{dx} = \frac{1}{x} \))

Applying the product rule:
\[ \frac{dy}{dx} = x \cdot \frac{1}{x} + \ln x \cdot 1 \]
This simplifies to:
\[ \frac{dy}{dx} = 1 + \ln x \]

### Step 2: Analyze the derivative
Now we have \( y = 1 + \ln x \). We need to analyze this function to understand its behavior.

1. **Domain**: The function \( \ln x \) is defined for \( x > 0 \).
2. **Intercept**: To find where \( y = 0 \):
\[ 1 + \ln x = 0 \implies \ln x = -1 \implies x = e^{-1} = \frac{1}{e} \]
This means the graph intersects the x-axis at \( x = \frac{1}{e} \).

### Step 3: Behavior of the function
- As \( x \to 0^+ \), \( \ln x \to -\infty \) and hence \( y \to -\infty \).
- At \( x = 1 \), \( \ln 1 = 0 \) so \( y = 1 \).
- As \( x \to \infty \), \( \ln x \to \infty \) and hence \( y \to \infty \).

### Step 4: Sketch the graph
From the analysis:
- The graph starts from \( -\infty \) as \( x \) approaches \( 0 \).
- It crosses the x-axis at \( x = \frac{1}{e} \).
- It passes through the point \( (1, 1) \).
- The graph approaches \( \infty \) as \( x \) increases.

Thus, the sketch of the graph of \( y = 1 + \ln x \) will look like this:

1. It will start from the bottom left (negative infinity).
2. It will cross the x-axis at \( x = \frac{1}{e} \).
3. It will rise and pass through the point \( (1, 1) \).
4. It will continue to rise towards infinity as \( x \) increases.

### Final Answer
The sketch graph of \( y = \frac{d}{dx}(x \ln x) = 1 + \ln x \) is a curve that starts from \( -\infty \), crosses the x-axis at \( x = \frac{1}{e} \), passes through \( (1, 1) \), and rises towards \( +\infty \).