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

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Answer 1 · 2017-06-28T14:29:28

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) #