Find the derivatives of the following :
`sinxcosx`

Answer

To find the derivative of the function \( y = \sin x \cos x \), 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 \( y' \) is given by:

\[ y' = u'v + uv' \]

### Step-by-step Solution:

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

2. **Differentiate Each Function**:
- The derivative of \( u \) (which is \( \sin x \)) is:
\[ u' = \cos x \]
- The derivative of \( v \) (which is \( \cos x \)) is:
\[ v' = -\sin x \]

3. **Apply the Product Rule**:
Now, apply the product rule:
\[ y' = u'v + uv' \]
Substituting the values of \( u, v, u', \) and \( v' \):
\[ y' = (\cos x)(\cos x) + (\sin x)(-\sin x) \]

4. **Simplify the Expression**:
This simplifies to:
\[ y' = \cos^2 x - \sin^2 x \]

5. **Use the Trigonometric Identity**:
We can recognize that \( \cos^2 x - \sin^2 x \) is a known trigonometric identity:
\[ \cos^2 x - \sin^2 x = \cos(2x) \]

6. **Final Result**:
Therefore, the derivative of \( y = \sin x \cos x \) is:
\[ y' = \cos(2x) \]