# How do you find the first and second derivative of #ln[lnx^2+1)]#?

# d/dx ln \ ln(x^2+1) \ = (2x)/((x^2+1)ln(x^2+1)) #

# d^2/(dx^2) ln \ ln(x^2+1) = ( 2((1-x^2)ln(x^2+1) - 2x^2)) / ((x^2+1)^2ln^2(x^2+1)) #

Suppose we let:

Using the chain rule we have:

To find the second derivative we will need to apply the quotient rule:

And using the product rule, combined with the chain rulewe have:

Injecting this result into our earlier findings we get:

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To find the first derivative, use the chain rule and the fact that the derivative of ln(u) is u'/u, where u is a function of x. For the second derivative, apply the chain rule and the derivative of 1/u, which is -u'/u^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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