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

Answer

To differentiate the function \( y = x^{\sin x + \cos x} \) with respect to \( x \), we will use logarithmic differentiation. Here are the steps:

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

### Step 2: Apply the logarithmic identity
Using the property of logarithms that states \( \ln(a^b) = b \ln a \), we can simplify the right side:
\[ \ln y = (\sin x + \cos x) \ln x \]

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

### Step 4: Differentiate the components
Now we differentiate each component:
- The derivative of \( \sin x + \cos x \) is \( \cos x - \sin x \).
- The derivative of \( \ln x \) is \( \frac{1}{x} \).

Substituting these derivatives back into the equation gives:
\[ \frac{1}{y} \frac{dy}{dx} = (\cos x - \sin x) \ln x + (\sin x + \cos x) \cdot \frac{1}{x} \]

### Step 5: Solve for \( \frac{dy}{dx} \)
Now, we can solve for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = y \left[ (\cos x - \sin x) \ln x + \frac{\sin x + \cos x}{x} \right] \]

### Step 6: Substitute back for \( y \)
Recall that \( y = x^{\sin x + \cos x} \):
\[ \frac{dy}{dx} = x^{\sin x + \cos x} \left[ (\cos x - \sin x) \ln x + \frac{\sin x + \cos x}{x} \right] \]

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

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