Find `(dy)/(dx)" when "y=(x+cosx)/(tanx)`.

Answer

To find \(\frac{dy}{dx}\) when \(y = \frac{x + \cos x}{\tan x}\), we will use the quotient rule for differentiation. The quotient rule states that if \(y = \frac{u}{v}\), then

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

where \(u = x + \cos x\) and \(v = \tan x\).

### Step 1: Identify \(u\) and \(v\)
Let:
- \(u = x + \cos x\)
- \(v = \tan x\)

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

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

2. Differentiate \(v\):
\[ \frac{dv}{dx} = \frac{d}{dx}(\tan x) = \sec^2 x \]

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

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

### Step 4: Simplify the expression
Now, we will simplify the expression:

1. Expand the numerator:
\[ \tan x (1 - \sin x) = \tan x - \tan x \sin x \]
Thus, the numerator becomes:
\[ \tan x - \tan x \sin x - (x + \cos x) \sec^2 x \]

2. Combine terms:
The final expression for \(\frac{dy}{dx}\) is:
\[ \frac{dy}{dx} = \frac{\tan x - \tan x \sin x - (x + \cos x) \sec^2 x}{\tan^2 x} \]

### Final Answer
Thus, the derivative \(\frac{dy}{dx}\) is:
\[ \frac{dy}{dx} = \frac{\tan x - \tan x \sin x - (x + \cos x) \sec^2 x}{\tan^2 x} \]

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