How do you find the first and second derivative of #y=ln(lnx^2)#?
The First Derivative is:
# dy/dx = 1/(xlnx) #
The Second Derivative is:
# (d^2y)/(dx^2) =  (1 + lnx)/(x^2ln^2x) #
We have:
Using the law of logarithms we can write this as:
If we write this as:
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To find the first and second derivatives of ( y = \ln(\ln(x^2)) ):

First derivative: [ \frac{dy}{dx} = \frac{1}{\ln(x^2)} \cdot \frac{d}{dx}(\ln(x^2)) = \frac{1}{\ln(x^2)} \cdot \frac{1}{x^2} \cdot 2x ]

Simplify the expression: [ \frac{dy}{dx} = \frac{2}{x\ln(x^2)} ]

Second derivative: [ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{2}{x\ln(x^2)}\right) = \frac{2}{x^2\ln(x^2)} + \frac{4}{x^2(\ln(x^2))^2} ]
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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