Question

√x (x²-2x+6)
Using the product rule.
Show each step

Answer 1

To differentiate \( y = \sqrt{x} (x^2 - 2x + 6) \) using the product rule, we will follow these steps:

1. **Identify the functions**:
Let
\[
u = \sqrt{x} \quad \text{and} \quad v = x^2 - 2x + 6
\]

2. **Differentiate \( u \) and \( v \)**:
- To differentiate \( u = \sqrt{x} = x^{1/2} \):
\[
u' = \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}}
\]
- To differentiate \( v = x^2 - 2x + 6 \):
\[
v' = 2x - 2
\]

3. **Apply the product rule**:
The product rule states:
\[
(uv)' = u'v + uv'
\]
Therefore, we can substitute \( u \), \( v \), \( u' \), and \( v' \):
\[
y' = u'v + uv'
\]
Substituting in the expressions we found:
\[
y' = \left(\frac{1}{2\sqrt{x}}\right)(x^2 - 2x + 6) + \sqrt{x}(2x - 2)
\]

4. **Simplify the expression**:
- The first term becomes:
\[
\frac{1}{2\sqrt{x}}(x^2 - 2x + 6)
\]
- The second term becomes:
\[
\sqrt{x}(2x - 2) = 2x\sqrt{x} - 2\sqrt{x}
\]

Putting these together, we have:
\[
y' = \frac{x^2 - 2x + 6}{2\sqrt{x}} + 2x\sqrt{x} - 2\sqrt{x}
\]

5. **Combine fractions** (if necessary) to simplify further:
To combine these terms, we can multiply \( 2x\sqrt{x} - 2\sqrt{x} \) by \( \frac{2\sqrt{x}}{2\sqrt{x}} \) to get a common denominator:
\[
y' = \frac{x^2 - 2x + 6}{2\sqrt{x}} + \frac{(2x\sqrt{x} - 2\sqrt{x})2\sqrt{x}}{2\sqrt{x}}
\]
This gives:
\[
y' = \frac{x^2 - 2x + 6 + 4x^2 - 4x}{2\sqrt{x}} = \frac{(x^2 + 4x^2) - (2x + 4x) + 6}{2\sqrt{x}}
\]
Combining the like terms:
\[
y' = \frac{5x^2 - 6x + 6}{2\sqrt{x}}
\]

Thus, the derivative of \( y = \sqrt{x} (x^2 - 2x + 6) \) is:
\[
y' = \frac{5x^2 - 6x + 6}{2\sqrt{x}}
\]