# How do you use the product rule to differentiate #g(s)=sqrts(4-s^2)#?

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To differentiate ( g(s) = \sqrt{4 - s^2} ) using the product rule, you first recognize it as the product of two functions: ( f(s) = \sqrt{4} ) and ( h(s) = \sqrt{4 - s^2} ). Then, apply the product rule ( (f \cdot h)' = f' \cdot h + f \cdot h' ) where ( f' ) is the derivative of ( f ) and ( h' ) is the derivative of ( h ). Differentiate each function separately and then apply the product rule.

( f(s) = \sqrt{4} ) is a constant function, so its derivative ( f'(s) ) is zero.

( h(s) = \sqrt{4 - s^2} ) can be differentiated using the chain rule: ( h'(s) = \frac{-s}{\sqrt{4 - s^2}} ).

Now, apply the product rule: [ g'(s) = f'(s) \cdot h(s) + f(s) \cdot h'(s) ] [ g'(s) = 0 \cdot \sqrt{4 - s^2} + \sqrt{4} \cdot \frac{-s}{\sqrt{4 - s^2}} ]

Simplify: [ g'(s) = \frac{-s\sqrt{4}}{\sqrt{4 - s^2}} ] [ g'(s) = \frac{-s \cdot 2}{\sqrt{4 - s^2}} ] [ g'(s) = \frac{-2s}{\sqrt{4 - s^2}} ]

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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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