Find `dy/dx if y= cosx/x`

Answer

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

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

where \(u = \cos x\) and \(v = x\).

### Step-by-step Solution:

1. **Identify \(u\) and \(v\)**:
- Let \(u = \cos x\)
- Let \(v = x\)

2. **Differentiate \(u\) and \(v\)**:
- The derivative of \(u\) with respect to \(x\) is:
\[ \frac{du}{dx} = -\sin x \]
- The derivative of \(v\) with respect to \(x\) is:
\[ \frac{dv}{dx} = 1 \]

3. **Apply the Quotient Rule**:
- Now, we can apply the quotient rule:
\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
Substituting \(u\), \(v\), \(\frac{du}{dx}\), and \(\frac{dv}{dx}\) into the formula:
\[ \frac{dy}{dx} = \frac{x(-\sin x) - \cos x(1)}{x^2} \]

4. **Simplify the Expression**:
- Simplifying the numerator:
\[ \frac{dy}{dx} = \frac{-x \sin x - \cos x}{x^2} \]
- This can be rewritten as:
\[ \frac{dy}{dx} = \frac{-x \sin x - \cos x}{x^2} \]

5. **Final Result**:
- Therefore, the derivative \(\frac{dy}{dx}\) is:
\[ \frac{dy}{dx} = \frac{-x \sin x - \cos x}{x^2} \]