If ` y = sin x^(@) " and " u = cos x " then " (dy)/(du)` is equal to

Options

A. ` - cosec x cos x`
B. ` pi/180 cosec "" (pix)/180 cos x `
C. ` - pi/180 cos ec x cos "" (pix)/180`
D. `pi/180 sin x ""(cos pi x)/180`

Answer

To solve the problem, we need to find the derivative \(\frac{dy}{du}\) given that \(y = \sin(x^\circ)\) and \(u = \cos(x)\).

### Step-by-Step Solution:

1. **Convert Degrees to Radians**:
Since \(y = \sin(x^\circ)\), we need to convert \(x\) from degrees to radians. The conversion from degrees to radians is done by multiplying by \(\frac{\pi}{180}\). Therefore, we can express \(y\) as:
\[ y = \sin\left(\frac{\pi x}{180}\right) \]

2. **Differentiate \(y\) with respect to \(x\)**:
We will use the chain rule to differentiate \(y\):
\[ \frac{dy}{dx} = \cos\left(\frac{\pi x}{180}\right) \cdot \frac{d}{dx}\left(\frac{\pi x}{180}\right) \]
The derivative of \(\frac{\pi x}{180}\) with respect to \(x\) is \(\frac{\pi}{180}\). Thus, we have:
\[ \frac{dy}{dx} = \cos\left(\frac{\pi x}{180}\right) \cdot \frac{\pi}{180} \]

3. **Differentiate \(u\) with respect to \(x\)**:
Now, we differentiate \(u = \cos(x)\):
\[ \frac{du}{dx} = -\sin(x) \]

4. **Apply the Chain Rule to find \(\frac{dy}{du}\)**:
By using the chain rule, we can express \(\frac{dy}{du}\) as:
\[ \frac{dy}{du} = \frac{dy}{dx} \cdot \frac{dx}{du} = \frac{dy}{dx} \cdot \frac{1}{\frac{du}{dx}} \]
Substituting the derivatives we found:
\[ \frac{dy}{du} = \frac{\cos\left(\frac{\pi x}{180}\right) \cdot \frac{\pi}{180}}{-\sin(x)} \]
Simplifying this gives:
\[ \frac{dy}{du} = -\frac{\pi \cos\left(\frac{\pi x}{180}\right)}{180 \sin(x)} \]

5. **Final Expression**:
Thus, the final expression for \(\frac{dy}{du}\) is:
\[ \frac{dy}{du} = -\frac{\pi}{180} \cdot \frac{\cos\left(\frac{\pi x}{180}\right)}{\sin(x)} \]