if `y = A sin(omega t - kx)`, then the value of `(dy)/(dx)` is

Options

A. `A cos (omega t - kx)`
B. `-A omega cos (omega t - kx)`
C. `AK cos (omega t - kx)`
D. `- AK cos (omega t - kx)`

Answer

To find the value of \(\frac{dy}{dx}\) for the function \(y = A \sin(\omega t - kx)\), we will differentiate \(y\) with respect to \(x\). Here are the steps to solve the problem:

### Step-by-Step Solution:

1. **Identify the function**: We have \(y = A \sin(\omega t - kx)\). In this function, \(A\) is a constant (amplitude), \(\omega\) is the angular frequency, \(t\) is time, \(k\) is the wave number, and \(x\) is the displacement.

2. **Differentiate with respect to \(x\)**: To find \(\frac{dy}{dx}\), we need to differentiate \(y\) with respect to \(x\). Since \(A\) and \(\omega t\) are constants with respect to \(x\), we treat them as such during differentiation.

3. **Apply the chain rule**: The derivative of \(\sin(u)\) with respect to \(u\) is \(\cos(u)\), where \(u = \omega t - kx\). Therefore, we have:
\[ \frac{dy}{dx} = A \cdot \cos(\omega t - kx) \cdot \frac{d}{dx}(\omega t - kx) \]

4. **Differentiate the inner function**: Now we differentiate the inner function \(\omega t - kx\) with respect to \(x\):
- The derivative of \(\omega t\) with respect to \(x\) is \(0\) (since \(\omega t\) is constant with respect to \(x\)).
- The derivative of \(-kx\) with respect to \(x\) is \(-k\).

5. **Combine the results**: Plugging this back into our equation gives:
\[ \frac{dy}{dx} = A \cdot \cos(\omega t - kx) \cdot (-k) \]
Simplifying this, we get:
\[ \frac{dy}{dx} = -A k \cos(\omega t - kx) \]

### Final Answer:
\[ \frac{dy}{dx} = -A k \cos(\omega t - kx) \]