What is the derivative of #ln(1-x^2)^(1/2)#?

Answer 1

I found: #f'(x)=-x/(1-x^2)#

I would write it as: #f(x)=ln[(1-x^2)^(1/2)]=1/2ln(1-x^2)# using one of the properties of logs. Then derive using the Chain Rule for the log and its argument as: #f'(x)=1/2*1/(1-x^2)*(-2x)=-x/(1-x^2)#
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Answer 2

To find the derivative of ( \ln(\sqrt{1-x^2}) ), we'll use the chain rule.

Let ( u = \sqrt{1-x^2} ), then ( \ln(u) = \ln(\sqrt{1-x^2}) ).

Now, we'll find ( \frac{du}{dx} ) and use it to differentiate ( \ln(u) ) with respect to ( x ).

[ u = \sqrt{1-x^2} ] [ \frac{du}{dx} = \frac{d}{dx} \sqrt{1-x^2} ]

Using the chain rule for differentiation of ( \sqrt{1-x^2} ), we get:

[ \frac{du}{dx} = \frac{d}{dx} (1-x^2)^{1/2} = -\frac{1}{2}(1-x^2)^{-1/2} \cdot (-2x) = \frac{x}{\sqrt{1-x^2}} ]

Now, using the chain rule, we have:

[ \frac{d}{dx} \ln(u) = \frac{1}{u} \cdot \frac{du}{dx} ]

Substituting the expression for ( \frac{du}{dx} ), we get:

[ \frac{d}{dx} \ln(u) = \frac{1}{\sqrt{1-x^2}} \cdot \frac{x}{\sqrt{1-x^2}} = \frac{x}{1-x^2} ]

Therefore, the derivative of ( \ln(\sqrt{1-x^2}) ) is ( \frac{x}{1-x^2} ).

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Answer from HIX Tutor

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.

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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