If ` y= ( log sin x) (logx ) ,then (dy)/(dx) `

Options

A. ` (log (sin x))/( x) +(log x )(cotx) `
B. ` (log (sin x))/( x) -(log x )(cotx) `
C. ` (log (sin x))/( x)+(log x )(tan x) `
D. ` (log (sin x))/( x)-(log x )(tan x) `

Answer

To find the derivative of the function \( y = (\log(\sin x)) (\log x) \), we will use the product rule of differentiation. The product rule states that if you have a function that is the product of two functions, say \( u \) and \( v \), then the derivative of their product is given by:

\[ \frac{d(uv)}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]

### Step-by-step Solution:

1. **Identify the functions**:
Let \( u = \log(\sin x) \) and \( v = \log x \).

2. **Differentiate \( u \)**:
To find \( \frac{du}{dx} \):
\[ u = \log(\sin x) \]
Using the chain rule, we have:
\[ \frac{du}{dx} = \frac{1}{\sin x} \cdot \frac{d(\sin x)}{dx} = \frac{1}{\sin x} \cdot \cos x = \cot x \]

3. **Differentiate \( v \)**:
To find \( \frac{dv}{dx} \):
\[ v = \log x \]
The derivative is:
\[ \frac{dv}{dx} = \frac{1}{x} \]

4. **Apply the product rule**:
Now, apply the product rule:
\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]
Substituting the values we found:
\[ \frac{dy}{dx} = \log(\sin x) \cdot \frac{1}{x} + \log x \cdot \cot x \]

5. **Final result**:
Therefore, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = \frac{\log(\sin x)}{x} + \log x \cdot \cot x \]