If ` y=log (sin x +cos x ) ,then (dy)/(dx) `

Options

A. ` (xsin x)/( xsin x+cos x ) `
B. ` (-xsin x)/( xsin x+cos x ) `
C. ` (xcos x)/( xsin x+cos x ) `
D. ` (-xcos x)/( xsin x+cos x ) `

Answer

To find the derivative \( \frac{dy}{dx} \) of the function \( y = \log(\sin x + \cos x) \), we will follow these steps:

### Step 1: Differentiate using the chain rule
The derivative of \( y = \log(f(x)) \) is given by:
\[ \frac{dy}{dx} = \frac{1}{f(x)} \cdot \frac{df}{dx} \]
where \( f(x) = \sin x + \cos x \).

### Step 2: Find \( f'(x) \)
Now we need to differentiate \( f(x) = \sin x + \cos x \):
\[ f'(x) = \frac{d}{dx}(\sin x) + \frac{d}{dx}(\cos x) = \cos x - \sin x \]

### Step 3: Substitute \( f(x) \) and \( f'(x) \) into the derivative formula
Now substituting \( f(x) \) and \( f'(x) \) into the derivative formula:
\[ \frac{dy}{dx} = \frac{1}{\sin x + \cos x} \cdot (\cos x - \sin x) \]

### Step 4: Simplify the expression
Thus, we can write:
\[ \frac{dy}{dx} = \frac{\cos x - \sin x}{\sin x + \cos x} \]

### Step 5: Further simplification (if necessary)
We can also express \( \cos x - \sin x \) using trigonometric identities, but for the purpose of this problem, the expression is already in a simplified form.

### Final Answer
The derivative is:
\[ \frac{dy}{dx} = \frac{\cos x - \sin x}{\sin x + \cos x} \]

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