if `y=(sin(x+9))/(cosx)`, then `(dy)/(dx)` at x=0 is equal to

Options

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

Explanation

To find the derivative of the function \( y = \frac{\sin(x + 9)}{\cos x} \) and evaluate it at \( x = 0 \), we can follow these steps:

### Step 1: Identify the function
We have:
\[ y = \frac{\sin(x + 9)}{\cos x} \]

### Step 2: Apply the Quotient Rule
The Quotient Rule states that if \( y = \frac{u}{v} \), then:
\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
Here, \( u = \sin(x + 9) \) and \( v = \cos x \).

### Step 3: Differentiate \( u \) and \( v \)
We need to find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \):
- For \( u = \sin(x + 9) \):
\[ \frac{du}{dx} = \cos(x + 9) \]
- For \( v = \cos x \):
\[ \frac{dv}{dx} = -\sin x \]

### Step 4: Substitute into the Quotient Rule
Now substitute \( u \), \( v \), \( \frac{du}{dx} \), and \( \frac{dv}{dx} \) into 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 5: Use the Angle Addition Formula
Using the angle addition formula for cosine:
\[ \cos a \cos b + \sin a \sin b = \cos(a - b) \]
We can rewrite the numerator:
\[ \cos x \cdot \cos(x + 9) + \sin(x + 9) \cdot \sin x = \cos((x + 9) - x) = \cos(9) \]
Thus, we have:
\[ \frac{dy}{dx} = \frac{\cos(9)}{\cos^2 x} \]

### Step 6: Evaluate at \( x = 0 \)
Now we evaluate \( \frac{dy}{dx} \) at \( x = 0 \):
\[ \frac{dy}{dx} \bigg|_{x=0} = \frac{\cos(9)}{\cos^2(0)} \]
Since \( \cos(0) = 1 \), we have:
\[ \frac{dy}{dx} \bigg|_{x=0} = \frac{\cos(9)}{1^2} = \cos(9) \]

### Final Answer
Thus, the value of \( \frac{dy}{dx} \) at \( x = 0 \) is:
\[ \cos(9) \]