Find `(dy)/(dx)` when :
`y=(cosx)/(1+sinx)`

Answer

To find \(\frac{dy}{dx}\) for the function \(y = \frac{\cos x}{1 + \sin x}\), we will use the quotient rule of differentiation. The quotient rule states that if you have a function in the form \(y = \frac{a}{b}\), then:

\[ \frac{dy}{dx} = \frac{b \frac{da}{dx} - a \frac{db}{dx}}{b^2} \]

where \(a = \cos x\) and \(b = 1 + \sin x\).

### Step 1: Identify \(a\) and \(b\)
Let:
- \(a = \cos x\)
- \(b = 1 + \sin x\)

### Step 2: Differentiate \(a\) and \(b\)
Now we need to find \(\frac{da}{dx}\) and \(\frac{db}{dx}\):
- \(\frac{da}{dx} = -\sin x\) (derivative of \(\cos x\))
- \(\frac{db}{dx} = \cos x\) (derivative of \(1 + \sin x\))

### Step 3: Apply the Quotient Rule
Substituting \(a\), \(b\), \(\frac{da}{dx}\), and \(\frac{db}{dx}\) into the quotient rule formula:

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

### Step 4: Simplify the Numerator
Now we simplify the numerator:

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

Using the Pythagorean identity \(\sin^2 x + \cos^2 x = 1\), we can replace \(\cos^2 x\):

\[ = \frac{-\sin x - \sin^2 x - (1 - \sin^2 x)}{(1 + \sin x)^2} \]

This simplifies to:

\[ = \frac{-\sin x - 1}{(1 + \sin x)^2} \]

### Step 5: Final Expression
Thus, we have:

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

### Step 6: Factor out the negative sign (optional)
Factoring out the negative sign gives:

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

This can be simplified to:

\[ \frac{dy}{dx} = -\frac{1}{1 + \sin x} \]

### Final Answer
Therefore, the derivative \(\frac{dy}{dx}\) is:

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

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