# How do you find the second derivative of #y= ln(1-x^2)^(1/2) #?

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To find the second derivative of ( y = \ln((1-x^2)^{1/2}) ), follow these steps:

- Find the first derivative of ( y ) using the chain rule.
- Differentiate the first derivative obtained in step 1 with respect to ( x ) to find the second derivative.

Let's proceed:

Given function: ( y = \ln((1-x^2)^{1/2}) )

- Find the first derivative: [ \frac{dy}{dx} = \frac{1}{(1-x^2)^{1/2}} \cdot \frac{d}{dx}(1-x^2)^{1/2} ]

Using the chain rule, differentiate ( (1-x^2)^{1/2} ) with respect to ( x ): [ \frac{d}{dx}(1-x^2)^{1/2} = \frac{1}{2}(1-x^2)^{-1/2} \cdot (-2x) ] [ = -\frac{x}{(1-x^2)^{1/2}} ]

So, the first derivative ( \frac{dy}{dx} ) is: [ \frac{dy}{dx} = -\frac{x}{(1-x^2)} ]

- Now, differentiate ( \frac{dy}{dx} ) with respect to ( x ) to find the second derivative: [ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(-\frac{x}{(1-x^2)}\right) ]

Using the quotient rule: [ \frac{d^2y}{dx^2} = \frac{(1-x^2)(-1) - (-x)(-2x)}{(1-x^2)^2} ] [ = \frac{-1 + x^2 + 2x^2}{(1-x^2)^2} ] [ = \frac{3x^2 - 1}{(1-x^2)^2} ]

Therefore, the second derivative ( \frac{d^2y}{dx^2} ) is: [ \frac{d^2y}{dx^2} = \frac{3x^2 - 1}{(1-x^2)^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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