# What is the second derivative of #y=x*sqrt(16-x^2)#?

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To find the second derivative of ( y = x \sqrt{16 - x^2} ), we first need to find the first derivative, and then differentiate again.

The first derivative, ( y' ), is calculated using the product rule: [ y' = (x)' \cdot \sqrt{16 - x^2} + x \cdot (\sqrt{16 - x^2})' ] [ y' = 1 \cdot \sqrt{16 - x^2} + x \cdot \frac{1}{2\sqrt{16 - x^2}} \cdot (-2x) ] [ y' = \sqrt{16 - x^2} - \frac{x^2}{\sqrt{16 - x^2}} ]

Now, we'll differentiate ( y' ) to find the second derivative, ( y'' ): [ y'' = \left( \sqrt{16 - x^2} \right)' - \left( \frac{x^2}{\sqrt{16 - x^2}} \right)' ] [ y'' = \frac{1}{2\sqrt{16 - x^2}} \cdot (-2x) - \frac{\sqrt{16 - x^2} \cdot 2x}{(16 - x^2)^{3/2}} ] [ y'' = -\frac{x}{\sqrt{16 - x^2}} - \frac{2x^2}{(16 - x^2)^{3/2}} ]

So, the second derivative of ( y = x \sqrt{16 - x^2} ) is ( y'' = -\frac{x}{\sqrt{16 - x^2}} - \frac{2x^2}{(16 - x^2)^{3/2}} ).

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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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