Differentiate the following w.r.t. x :
`(logx)^(x)+(x)^(cosx)`

Answer

To differentiate the function \( y = (\log x)^x + x^{\cos x} \) with respect to \( x \), we will follow these steps:

### Step 1: Define the function
Let:
\[ y = (\log x)^x + x^{\cos x} \]

### Step 2: Differentiate the first term \( u = (\log x)^x \)
To differentiate \( u = (\log x)^x \), we can use logarithmic differentiation:
1. Take the natural logarithm of both sides:
\[ \log u = x \log(\log x) \]
2. Differentiate both sides with respect to \( x \):
\[ \frac{1}{u} \frac{du}{dx} = \log(\log x) + x \cdot \frac{1}{\log x} \cdot \frac{1}{x} \]
Simplifying gives:
\[ \frac{1}{u} \frac{du}{dx} = \log(\log x) + \frac{1}{\log x} \]
3. Multiply by \( u \) to solve for \( \frac{du}{dx} \):
\[ \frac{du}{dx} = u \left( \log(\log x) + \frac{1}{\log x} \right) \]
Substituting back \( u = (\log x)^x \):
\[ \frac{du}{dx} = (\log x)^x \left( \log(\log x) + \frac{1}{\log x} \right) \]

### Step 3: Differentiate the second term \( v = x^{\cos x} \)
For \( v = x^{\cos x} \), we again use logarithmic differentiation:
1. Take the natural logarithm:
\[ \log v = \cos x \log x \]
2. Differentiate both sides:
\[ \frac{1}{v} \frac{dv}{dx} = -\sin x \log x + \cos x \cdot \frac{1}{x} \]
3. Multiply by \( v \):
\[ \frac{dv}{dx} = v \left( -\sin x \log x + \frac{\cos x}{x} \right) \]
Substituting back \( v = x^{\cos x} \):
\[ \frac{dv}{dx} = x^{\cos x} \left( -\sin x \log x + \frac{\cos x}{x} \right) \]

### Step 4: Combine the derivatives
Now, we can find \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \]
Substituting the expressions we found:
\[ \frac{dy}{dx} = (\log x)^x \left( \log(\log x) + \frac{1}{\log x} \right) + x^{\cos x} \left( -\sin x \log x + \frac{\cos x}{x} \right) \]

### Final Answer
Thus, the derivative of the function \( y \) with respect to \( x \) is:
\[ \frac{dy}{dx} = (\log x)^x \left( \log(\log x) + \frac{1}{\log x} \right) + x^{\cos x} \left( -\sin x \log x + \frac{\cos x}{x} \right) \]