# How do you find the first and second derivative of #ln(x+sqrt((x^2)-1))#?

Recall that

The derivative of this expression will require use of the chain rule on our square root...

This is our du term. Our u is already given, so du/u is

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To find the first derivative of ( \ln(x + \sqrt{x^2 - 1}) ), use the chain rule and the derivative of the natural logarithm function:

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

Let ( u = x + \sqrt{x^2 - 1} ). Then:

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

Now, applying the chain rule:

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

For the second derivative, differentiate the expression we found for the first derivative:

[ \frac{d}{dx}\left[\frac{1}{x + \sqrt{x^2 - 1}} \cdot \left(1 + \frac{x}{\sqrt{x^2 - 1}}\right)\right] ]

After simplifying, you will get the second derivative.

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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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