Differentiate the following w.r.t. x :
`x^(x)-2^(sinx)`

Answer

To differentiate the function \( y = x^x - 2^{\sin x} \) with respect to \( x \), we will follow these steps:

### Step 1: Differentiate \( y = x^x \)

To differentiate \( x^x \), we will use logarithmic differentiation.

1. Start by taking the natural logarithm of both sides:
\[ \ln y = \ln(x^x) \]

2. Using the property of logarithms, we can simplify:
\[ \ln y = x \ln x \]

3. Now differentiate both sides with respect to \( x \):
\[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(x \ln x) \]

4. Apply the product rule on the right side:
\[ \frac{d}{dx}(x \ln x) = \ln x + 1 \]

5. Now, substitute back for \( y \):
\[ \frac{1}{y} \frac{dy}{dx} = \ln x + 1 \]

6. Multiply both sides by \( y \):
\[ \frac{dy}{dx} = y(\ln x + 1) = x^x(\ln x + 1) \]

### Step 2: Differentiate \( y = 2^{\sin x} \)

Now, we will differentiate \( 2^{\sin x} \):

1. Use the chain rule:
\[ \frac{d}{dx}(2^{\sin x}) = 2^{\sin x} \ln(2) \cdot \frac{d}{dx}(\sin x) \]

2. Since \( \frac{d}{dx}(\sin x) = \cos x \):
\[ \frac{d}{dx}(2^{\sin x}) = 2^{\sin x} \ln(2) \cos x \]

### Step 3: Combine the results

Now, we can combine the derivatives of both parts:

1. The derivative of \( y = x^x - 2^{\sin x} \):
\[ \frac{dy}{dx} = \frac{d}{dx}(x^x) - \frac{d}{dx}(2^{\sin x}) \]

2. Substitute the derivatives we found:
\[ \frac{dy}{dx} = x^x(\ln x + 1) - 2^{\sin x} \ln(2) \cos x \]

### Final Answer

Thus, the derivative of the function \( y = x^x - 2^{\sin x} \) with respect to \( x \) is:
\[ \frac{dy}{dx} = x^x(\ln x + 1) - 2^{\sin x} \ln(2) \cos x \]

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