If y =sin (sin x) and `(d^(2)y)/(dx^(2))+(dy)/(dx)` tan x + f(x) = 0, then find f(x).

Explanation

To solve the problem, we need to find \( f(x) \) given the equation:

\[ \frac{d^2y}{dx^2} + \frac{dy}{dx} \tan x + f(x) = 0 \]

where \( y = \sin(\sin x) \).

### Step 1: Find \( \frac{dy}{dx} \)

We start by differentiating \( y \) with respect to \( x \):

\[ y = \sin(\sin x) \]

Using the chain rule, we differentiate:

\[ \frac{dy}{dx} = \cos(\sin x) \cdot \frac{d}{dx}(\sin x) = \cos(\sin x) \cdot \cos x \]

### Step 2: Find \( \frac{d^2y}{dx^2} \)

Next, we differentiate \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} = \cos(\sin x) \cos x \]

Using the product rule:

Let \( u = \cos(\sin x) \) and \( v = \cos x \).

Then,

\[ \frac{d^2y}{dx^2} = \frac{du}{dx} v + u \frac{dv}{dx} \]

Calculating \( \frac{du}{dx} \):

\[ \frac{du}{dx} = -\sin(\sin x) \cdot \cos x \]

Calculating \( \frac{dv}{dx} \):

\[ \frac{dv}{dx} = -\sin x \]

Substituting back into the product rule:

\[ \frac{d^2y}{dx^2} = (-\sin(\sin x) \cdot \cos x) \cdot \cos x + \cos(\sin x) \cdot (-\sin x) \]

This simplifies to:

\[ \frac{d^2y}{dx^2} = -\sin(\sin x) \cos^2 x - \cos(\sin x) \sin x \]

### Step 3: Substitute into the original equation

Now we substitute \( \frac{dy}{dx} \) and \( \frac{d^2y}{dx^2} \) into the original equation:

\[ -\sin(\sin x) \cos^2 x - \cos(\sin x) \sin x + \left( \cos(\sin x) \cos x \right) \tan x + f(x) = 0 \]

### Step 4: Simplify the equation

Using \( \tan x = \frac{\sin x}{\cos x} \):

\[ -\sin(\sin x) \cos^2 x - \cos(\sin x) \sin x + \cos(\sin x) \cos x \cdot \frac{\sin x}{\cos x} + f(x) = 0 \]

This simplifies to:

\[ -\sin(\sin x) \cos^2 x - \cos(\sin x) \sin x + \cos(\sin x) \sin x + f(x) = 0 \]

The \( -\cos(\sin x) \sin x \) and \( +\cos(\sin x) \sin x \) cancel out:

\[ -\sin(\sin x) \cos^2 x + f(x) = 0 \]

### Step 5: Solve for \( f(x) \)

Rearranging gives:

\[ f(x) = \sin(\sin x) \cos^2 x \]

### Final Answer

Thus, the function \( f(x) \) is:

\[ \boxed{f(x) = \sin(\sin x) \cos^2 x} \]