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

##### 1 Answer

Jun 28, 2017

The First Derivative is:

# dy/dx = 1/(xlnx) #

The Second Derivative is:

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

#### Explanation:

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

# 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

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

# " " = -1/(xlnx)^2 \ (1 + lnx) #

# " " = - (1 + lnx)/(x^2ln^2x) #