# How do you differentiate #g(y) =xsqrtx # using the product rule?

For

When we use the product rule, we obtain

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To differentiate ( g(y) = x \sqrt{x} ) using the product rule:

- Identify the functions: ( f(x) = x ) and ( h(x) = \sqrt{x} ).
- Compute the derivatives of each function: ( f'(x) = 1 ) and ( h'(x) = \frac{1}{2\sqrt{x}} ).
- Apply the product rule: ( g'(x) = f(x)h'(x) + f'(x)h(x) ).
- Substitute the derivatives and functions: ( g'(x) = x \cdot \frac{1}{2\sqrt{x}} + 1 \cdot \sqrt{x} ).
- Simplify the expression: ( g'(x) = \frac{1}{2\sqrt{x}} \cdot x + \sqrt{x} ).
- Further simplify if needed: ( g'(x) = \frac{x}{2\sqrt{x}} + \sqrt{x} ).
- Finally, simplify the expression: ( g'(x) = \frac{x}{2\sqrt{x}} + \sqrt{x} = \frac{1}{2} \sqrt{x} + \sqrt{x} = \frac{3}{2} \sqrt{x} ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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