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 \(\frac{dy}{dx}\) of the function \(y = \frac{\sin(x + 9)}{\cos x}\) at \(x = 0\), we will apply the quotient rule of differentiation. The quotient rule states that if you have a function \(y = \frac{u}{v}\), then the derivative 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: Find \(\frac{du}{dx}\) and \(\frac{dv}{dx}\)
Now we need to find the derivatives of \(u\) and \(v\):
- \(\frac{du}{dx} = \cos(x + 9)\) (using the chain rule)
- \(\frac{dv}{dx} = -\sin x\)

### Step 3: Apply the Quotient Rule
Now, we can apply 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: Simplify the Numerator
Using the cosine addition formula, we can simplify 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 5: Evaluate at \(x = 0\)
Now we need to evaluate this derivative at \(x = 0\):

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

Since \(\cos(0) = 1\):

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

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

\[ \cos(9) \]