`((e^(x) + sinx)/(1+logx))`

Answer

To differentiate the function \( y = \frac{e^x + \sin x}{1 + \log x} \), we will use the quotient rule of differentiation. The quotient rule states that if you have a function in the form \( \frac{u}{v} \), then its derivative is given by:

\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

where \( u = e^x + \sin x \) and \( v = 1 + \log x \).

### Step 1: Identify \( u \) and \( v \)
Let:
- \( u = e^x + \sin x \)
- \( v = 1 + \log x \)

### Step 2: Differentiate \( u \) and \( v \)
Now we need to find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \).

1. **Differentiate \( u \)**:
\[ \frac{du}{dx} = \frac{d}{dx}(e^x) + \frac{d}{dx}(\sin x) = e^x + \cos x \]

2. **Differentiate \( v \)**:
\[ \frac{dv}{dx} = \frac{d}{dx}(1) + \frac{d}{dx}(\log x) = 0 + \frac{1}{x} = \frac{1}{x} \]

### Step 3: Apply the Quotient Rule
Now we can apply the quotient rule:

\[ \frac{dy}{dx} = \frac{(1 + \log x)(e^x + \cos x) - (e^x + \sin x)\left(\frac{1}{x}\right)}{(1 + \log x)^2} \]

### Step 4: Simplify the Expression
Now we simplify the numerator:

1. **Expand the first term**:
\[ (1 + \log x)(e^x + \cos x) = e^x + \cos x + e^x \log x + \cos x \log x \]

2. **Expand the second term**:
\[ (e^x + \sin x)\left(\frac{1}{x}\right) = \frac{e^x}{x} + \frac{\sin x}{x} \]

Putting it all together, the numerator becomes:

\[ e^x + \cos x + e^x \log x + \cos x \log x - \left(\frac{e^x}{x} + \frac{\sin x}{x}\right) \]

### Final Expression
Thus, the derivative is:

\[ \frac{dy}{dx} = \frac{e^x + \cos x + e^x \log x + \cos x \log x - \frac{e^x}{x} - \frac{\sin x}{x}}{(1 + \log x)^2} \]