The derivative of `ln(x+sinx)` with respect to `(x+cosx)` is

Options

A. `(1+cosx)/((x+sinx)(1-sinx))`
B. `(1-cosx)/((x+sinx)(1+sinx))`
C. `(1-cosx)/((x-sinx)(1+cosx))`
D. `(1+cosx)/((x-sinx)(1-cosx))`

Answer

To find the derivative of \( \ln(x + \sin x) \) with respect to \( x + \cos x \), we will use the chain rule. Let's denote:

- \( f(x) = \ln(x + \sin x) \)
- \( g(x) = x + \cos x \)

We need to compute \( \frac{d}{dg} f(x) \), which can be expressed using the chain rule as:

\[ \frac{d}{dg} f(x) = \frac{f'(x)}{g'(x)} \]

### Step 1: Differentiate \( f(x) \)

To differentiate \( f(x) = \ln(x + \sin x) \), we apply the chain rule:

\[ f'(x) = \frac{1}{x + \sin x} \cdot \frac{d}{dx}(x + \sin x) \]

Now, differentiate \( x + \sin x \):

\[ \frac{d}{dx}(x + \sin x) = 1 + \cos x \]

So, we can substitute this back into our expression for \( f'(x) \):

\[ f'(x) = \frac{1 + \cos x}{x + \sin x} \]

### Step 2: Differentiate \( g(x) \)

Next, we differentiate \( g(x) = x + \cos x \):

\[ g'(x) = \frac{d}{dx}(x + \cos x) = 1 - \sin x \]

### Step 3: Apply the chain rule

Now we can substitute \( f'(x) \) and \( g'(x) \) into our expression for \( \frac{d}{dg} f(x) \):

\[ \frac{d}{dg} f(x) = \frac{f'(x)}{g'(x)} = \frac{\frac{1 + \cos x}{x + \sin x}}{1 - \sin x} \]

### Step 4: Simplify the expression

This simplifies to:

\[ \frac{d}{dg} f(x) = \frac{1 + \cos x}{(x + \sin x)(1 - \sin x)} \]

### Final Answer

Thus, the derivative of \( \ln(x + \sin x) \) with respect to \( x + \cos x \) is:

\[ \frac{1 + \cos x}{(x + \sin x)(1 - \sin x)} \]

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