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

Answer 1

The First Derivative is:

# dy/dx = 1/(xlnx) #

The Second Derivative is:

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

We have:

# y = ln(lnx^2) #

Using the law of logarithms we can write this as:

# y = ln(2lnx) # # \ \ = ln2 + ln(lnx) #
Then Differentiating wrt #x# and applying the chain rule:
# dy/dx = 0 + 1/lnx * 1/x # # " " = 1/(xlnx) #

If we write this as:

# dy/dx = (xlnx)^(-1) #
Then we get the second derivative by differentiating again wrt #x# and applying the chain and product rule:
# (d^2y)/(dx^2) = (-1)(xlnx)^(-2)(x 1/x + 1.lnx) # # " " = -1/(xlnx)^2 \ (1 + lnx) # # " " = - (1 + lnx)/(x^2ln^2x) #
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Answer 2

To find the first and second derivatives of ( y = \ln(\ln(x^2)) ):

  1. 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 ]

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

  3. 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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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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