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

For the first, using chain rule, which states that

For the second, using quotient and product rules.

For the second derivative, we must recall quotient rule...

...and product rule (to derive the quotient):

Thus:

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To find the first and second derivatives of ( \ln(\ln(x^2)) ), we'll apply the chain rule and the derivative of the natural logarithm function.

First Derivative:

( \frac{d}{dx}[\ln(\ln(x^2))] )

( = \frac{1}{\ln(x^2)} \cdot \frac{1}{x^2} \cdot 2x )

( = \frac{2}{x \ln(x^2)} )

Second Derivative:

To find the second derivative, we differentiate the expression we obtained for the first derivative.

( \frac{d}{dx} \left[ \frac{2}{x \ln(x^2)} \right] )

Using the quotient rule:

( = \frac{2 \cdot (-x \ln(x^2))' - 2x \ln(x^2)'}{(x \ln(x^2))^2} )

( = \frac{2 \cdot (-1) \cdot \ln(x^2) - 2x \cdot \frac{1}{x^2} \cdot 2x}{(x \ln(x^2))^2} )

( = \frac{-2\ln(x^2) - 4}{(x \ln(x^2))^2} )

( = \frac{-2\ln(x^2) - 4}{(x \cdot 2\ln(x))^2} )

( = \frac{-2\ln(x^2) - 4}{(2x\ln(x))^2} )

( = \frac{-\ln(x^2) - 2}{(x\ln(x))^2} )

So, the second derivative of ( \ln(\ln(x^2)) ) is ( \frac{-\ln(x^2) - 2}{(x\ln(x))^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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