If `y=x^(x^(x)), "then find "(dy)/(dx)`.

Answer

To find the derivative of the function \( y = x^{x^{x}} \), we will follow these steps:

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

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

### Step 3: Differentiate both sides
Now, we differentiate both sides with respect to \( x \). We will use implicit differentiation on the left side and the product rule on the right side:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(x^{x} \ln x) \]

### Step 4: Differentiate the right-hand side
To differentiate \( x^{x} \ln x \), we first need to find the derivative of \( x^{x} \). Let \( z = x^{x} \). Taking the logarithm of both sides:
\[ \ln z = x \ln x \]
Differentiating both sides gives:
\[ \frac{1}{z} \frac{dz}{dx} = \ln x + 1 \]
Thus, we have:
\[ \frac{dz}{dx} = z(\ln x + 1) = x^{x}(\ln x + 1) \]

Now, we apply the product rule to \( x^{x} \ln x \):
\[ \frac{d}{dx}(x^{x} \ln x) = \frac{dz}{dx} \ln x + x^{x} \cdot \frac{1}{x} \]
Substituting \( \frac{dz}{dx} \):
\[ = x^{x}(\ln x + 1) \ln x + x^{x} \]
Factoring out \( x^{x} \):
\[ = x^{x} \left( (\ln x + 1) \ln x + 1 \right) \]

### Step 5: Substitute back into the derivative equation
Now substituting back into our earlier equation:
\[ \frac{1}{y} \frac{dy}{dx} = x^{x} \left( (\ln x + 1) \ln x + 1 \right) \]

### Step 6: Solve for \( \frac{dy}{dx} \)
To isolate \( \frac{dy}{dx} \), we multiply both sides by \( y \):
\[ \frac{dy}{dx} = y \cdot x^{x} \left( (\ln x + 1) \ln x + 1 \right) \]
Substituting back \( y = x^{x^{x}} \):
\[ \frac{dy}{dx} = x^{x^{x}} \cdot x^{x} \left( (\ln x + 1) \ln x + 1 \right) \]

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

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