Find the derivatives of the following :
`(x^(2)-5x+6) secx`

Answer

To find the derivative of the function \( y = (x^2 - 5x + 6) \sec x \), 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 is given by:

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

### Step-by-Step Solution:

1. **Identify the functions**:
Let \( u = x^2 - 5x + 6 \) and \( v = \sec x \).

2. **Differentiate \( u \)**:
\[ u' = \frac{d}{dx}(x^2 - 5x + 6) = 2x - 5 \]

3. **Differentiate \( v \)**:
The derivative of \( \sec x \) is \( \sec x \tan x \).
\[ v' = \frac{d}{dx}(\sec x) = \sec x \tan x \]

4. **Apply the product rule**:
Using the product rule:
\[ y' = u'v + uv' \]
Substitute \( u \), \( u' \), \( v \), and \( v' \):
\[ y' = (2x - 5) \sec x + (x^2 - 5x + 6)(\sec x \tan x) \]

5. **Factor out \( \sec x \)**:
We can factor out \( \sec x \) from both terms:
\[ y' = \sec x \left( (2x - 5) + (x^2 - 5x + 6) \tan x \right) \]

6. **Final expression**:
Thus, the derivative of the function is:
\[ y' = \sec x \left( (2x - 5) + (x^2 - 5x + 6) \tan x \right) \]