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

Answer

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

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

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

### Step-by-Step Solution:

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

2. **Differentiate \(u\) and \(v\)**:
- \(\frac{du}{dx} = \cos x\) (the derivative of \(\sin x\))
- \(\frac{dv}{dx} = 1\) (the derivative of \(x\))

3. **Apply the Quotient Rule**:
- According to the quotient rule:
\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
- Substitute \(u\), \(v\), \(\frac{du}{dx}\), and \(\frac{dv}{dx}\):
\[ \frac{dy}{dx} = \frac{x \cdot \cos x - \sin x \cdot 1}{x^2} \]

4. **Simplify the expression**:
- This simplifies to:
\[ \frac{dy}{dx} = \frac{x \cos x - \sin x}{x^2} \]

Thus, the derivative \(\frac{dy}{dx}\) of the function \(y = \frac{\sin x}{x}\) is:

\[ \frac{dy}{dx} = \frac{x \cos x - \sin x}{x^2} \]