If `y=x^(e^x) `, then ` (dy)/(dx)`=

Options

A. `y(log x+ e^x)`
B. `y log x (1/2 +e^x)`
C. `ye^x (log x +x)`
D. `ye^x (log x +(1)/(x))`

Answer

To find the derivative of the function \( y = x^{e^x} \), we can use logarithmic differentiation. Here’s a step-by-step solution:

### Step 1: Take the natural logarithm of both sides
We start by taking the natural logarithm of both sides of the equation:
\[ \ln y = \ln(x^{e^x}) \]

### Step 2: Simplify using logarithmic properties
Using the property of logarithms that states \( \ln(a^b) = b \ln a \), we can simplify the right-hand side:
\[ \ln y = e^x \ln x \]

### Step 3: Differentiate both sides
Now we differentiate both sides with respect to \( x \). For the left-hand side, we use the chain rule:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(e^x \ln x) \]

### Step 4: Apply the product rule on the right-hand side
To differentiate \( e^x \ln x \), we apply the product rule:
\[ \frac{d}{dx}(e^x \ln x) = e^x \cdot \frac{d}{dx}(\ln x) + \ln x \cdot \frac{d}{dx}(e^x) \]
Calculating the derivatives, we have:
\[ \frac{d}{dx}(\ln x) = \frac{1}{x} \quad \text{and} \quad \frac{d}{dx}(e^x) = e^x \]
Thus, substituting these derivatives back in gives:
\[ \frac{d}{dx}(e^x \ln x) = e^x \cdot \frac{1}{x} + \ln x \cdot e^x \]
This simplifies to:
\[ \frac{d}{dx}(e^x \ln x) = \frac{e^x}{x} + e^x \ln x \]

### Step 5: Substitute back into the equation
Now we can substitute this back into our differentiation equation:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{e^x}{x} + e^x \ln x \]

### Step 6: Solve for \( \frac{dy}{dx} \)
To isolate \( \frac{dy}{dx} \), we multiply both sides by \( y \):
\[ \frac{dy}{dx} = y \left( \frac{e^x}{x} + e^x \ln x \right) \]

### Step 7: Substitute back for \( y \)
Since we know \( y = x^{e^x} \), we substitute back:
\[ \frac{dy}{dx} = x^{e^x} \left( \frac{e^x}{x} + e^x \ln x \right) \]

### Step 8: Factor out \( e^x \)
Factoring \( e^x \) out gives:
\[ \frac{dy}{dx} = x^{e^x} e^x \left( \frac{1}{x} + \ln x \right) \]

### Final Result
Thus, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = e^x x^{e^x} \left( \frac{1}{x} + \ln x \right) \]

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