If `y=(sin(x+9))/(cosx)`, then `(dy)/(dx)` at` x = 0` is:

Options

A. `cos9`
B. `sin9`
C. 0
D. 1

Answer

To find the derivative of the function \( y = \frac{\sin(x + 9)}{\cos x} \) at \( x = 0 \), we will use the quotient rule for differentiation. The quotient rule states that if you have a function in the form \( y = \frac{u}{v} \), then the derivative \( \frac{dy}{dx} \) is given by:

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

where \( u = \sin(x + 9) \) and \( v = \cos x \).

### Step 1: Identify \( u \) and \( v \)
Let:
- \( u = \sin(x + 9) \)
- \( v = \cos x \)

### Step 2: Differentiate \( u \) and \( v \)
Now we need to find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \):
- \( \frac{du}{dx} = \cos(x + 9) \) (using the chain rule)
- \( \frac{dv}{dx} = -\sin x \)

### Step 3: Apply the Quotient Rule
Using the quotient rule:

\[ \frac{dy}{dx} = \frac{\cos x \cdot \cos(x + 9) - \sin(x + 9) \cdot (-\sin x)}{\cos^2 x} \]

This simplifies to:

\[ \frac{dy}{dx} = \frac{\cos x \cdot \cos(x + 9) + \sin(x + 9) \cdot \sin x}{\cos^2 x} \]

### Step 4: Evaluate at \( x = 0 \)
Now we need to evaluate \( \frac{dy}{dx} \) at \( x = 0 \):

1. Calculate \( \cos(0) = 1 \)
2. Calculate \( \sin(0) = 0 \)
3. Calculate \( \cos(0 + 9) = \cos(9) \)
4. Calculate \( \sin(0 + 9) = \sin(9) \)

Substituting these values into the derivative:

\[ \frac{dy}{dx} \bigg|_{x=0} = \frac{1 \cdot \cos(9) + \sin(9) \cdot 0}{1^2} \]

This simplifies to:

\[ \frac{dy}{dx} \bigg|_{x=0} = \cos(9) \]

### Final Answer
Thus, the value of \( \frac{dy}{dx} \) at \( x = 0 \) is:

\[ \frac{dy}{dx} \bigg|_{x=0} = \cos(9) \]