# How do you find the derivative of #g(x)=xsqrtx#?

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To find the derivative of ( g(x) = x \sqrt{x} ), you can use the product rule. The product rule states that if you have two functions, ( u(x) ) and ( v(x) ), their derivative with respect to ( x ) is given by ( u'(x)v(x) + u(x)v'(x) ). In this case, let ( u(x) = x ) and ( v(x) = \sqrt{x} ). Then, differentiate each function:

( u'(x) = 1 ) (derivative of ( x ) with respect to ( x )) ( v'(x) = \frac{1}{2\sqrt{x}} ) (derivative of ( \sqrt{x} ) with respect to ( x ))

Now, apply the product rule:

( g'(x) = u'(x)v(x) + u(x)v'(x) ) ( = (1)(\sqrt{x}) + (x)\left(\frac{1}{2\sqrt{x}}\right) ) ( = \sqrt{x} + \frac{x}{2\sqrt{x}} )

Finally, simplify the expression:

( g'(x) = \sqrt{x} + \frac{x}{2\sqrt{x}} ) ( = \sqrt{x} + \frac{1}{2}\sqrt{x} ) ( = \frac{3}{2}\sqrt{x} )

Therefore, the derivative of ( g(x) = x \sqrt{x} ) is ( g'(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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