To differentiate \( y = \sqrt{x} (x^2 - 2x + 6) \) using the product rule, we will follow these steps:
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Identify the functions: Let \[ u = \sqrt{x} \quad \text{and} \quad v = x^2 - 2x + 6 \]
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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 \( v = x^2 - 2x + 6 \): \[ v' = 2x - 2 \]
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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: \[ y' = \left(\frac{1}{2\sqrt{x}}\right)(x^2 - 2x + 6) + \sqrt{x}(2x - 2) \]
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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: \[ y' = \frac{x^2 - 2x + 6}{2\sqrt{x}} + 2x\sqrt{x} - 2\sqrt{x} \]
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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: \[ y' = \frac{5x^2 - 6x + 6}{2\sqrt{x}} \]