Find `(dy)/(dx)`, when:
`y=sin(x^(x))`

Explanation

To find \(\frac{dy}{dx}\) for the function \(y = \sin(x^x)\), we will use the chain rule and implicit differentiation. Here’s a step-by-step solution:

### Step 1: Differentiate the outer function
We start by differentiating \(y = \sin(u)\) where \(u = x^x\). The derivative of \(\sin(u)\) with respect to \(u\) is:
\[ \frac{dy}{du} = \cos(u) \]

### Step 2: Differentiate the inner function
Next, we need to differentiate \(u = x^x\). To do this, we will use logarithmic differentiation. Taking the natural logarithm of both sides gives:
\[ \ln(u) = \ln(x^x) = x \ln(x) \]

### Step 3: Differentiate using implicit differentiation
Now, we differentiate both sides with respect to \(x\):
\[ \frac{1}{u} \frac{du}{dx} = \frac{d}{dx}(x \ln(x)) \]
Using the product rule on the right side:
\[ \frac{d}{dx}(x \ln(x)) = \ln(x) + 1 \]
Thus, we have:
\[ \frac{1}{u} \frac{du}{dx} = \ln(x) + 1 \]
Multiplying both sides by \(u\) gives:
\[ \frac{du}{dx} = u(\ln(x) + 1) \]

### Step 4: Substitute back for \(u\)
Since \(u = x^x\), we substitute back:
\[ \frac{du}{dx} = x^x(\ln(x) + 1) \]

### Step 5: Combine using the chain rule
Now we can find \(\frac{dy}{dx}\) using the chain rule:
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \cos(u) \cdot \frac{du}{dx} \]
Substituting \(u = x^x\) and \(\frac{du}{dx}\):
\[ \frac{dy}{dx} = \cos(x^x) \cdot x^x(\ln(x) + 1) \]

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

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