If ` sin ( x+y) +cos (x+y) =log (x+y) ,then (d^(2)y)/(dx^(2))=`

Options

A. ` (-y)/( x)`
B. ` -1`
C. ` 0`
D. ` 1`

Answer

To solve the problem, we need to differentiate the given equation twice. Let's go through the steps systematically.

### Step 1: Differentiate the given equation

We start with the equation:

\[ \sin(x+y) + \cos(x+y) = \log(x+y) \]

We will differentiate both sides with respect to \(x\).

### Step 2: Apply the chain rule

Differentiating the left side:

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

Using the chain rule:

\[ \cos(x+y) \left(1 + \frac{dy}{dx}\right) - \sin(x+y) \left(1 + \frac{dy}{dx}\right) \]

Now differentiating the right side:

\[ \frac{d}{dx}[\log(x+y)] = \frac{1}{x+y} \left(1 + \frac{dy}{dx}\right) \]

### Step 3: Set the derivatives equal

Now we set the derivatives equal to each other:

\[ \cos(x+y) \left(1 + \frac{dy}{dx}\right) - \sin(x+y) \left(1 + \frac{dy}{dx}\right) = \frac{1}{x+y} \left(1 + \frac{dy}{dx}\right) \]

### Step 4: Factor out \(1 + \frac{dy}{dx}\)

We can factor out \(1 + \frac{dy}{dx}\) from both sides:

\[ (1 + \frac{dy}{dx}) \left(\cos(x+y) - \sin(x+y)\right) = \frac{1}{x+y} (1 + \frac{dy}{dx}) \]

### Step 5: Solve for \(\frac{dy}{dx}\)

Assuming \(1 + \frac{dy}{dx} \neq 0\), we can divide both sides by \(1 + \frac{dy}{dx}\):

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

### Step 6: Find \(\frac{dy}{dx}\)

Now we can rearrange the equation to find \(\frac{dy}{dx}\):

\[ 1 + \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -1 \]

### Step 7: Differentiate again to find \(\frac{d^2y}{dx^2}\)

Now we differentiate \(\frac{dy}{dx}\) again:

\[ \frac{d^2y}{dx^2} = \frac{d}{dx}[-1] = 0 \]

### Final Answer

Thus, we have:

\[ \frac{d^2y}{dx^2} = 0 \]