How do you find the derivative of #y = 9 ln(4 ln x)#?

1 Answer
Dec 26, 2017

Apply the chain rule.

Explanation:

Begin with the derivative of the outermost function, #ln#. We know that the derivative of the natural log is just one over the argument of the function. In this case, the argument is #4ln(x)#, so we begin with #1/(4ln(x))#. Of course #9# is a constant, so we leave it alone.

#=>y'=9*1/(4ln(x))*"..."#

Now we multiply by the derivative of the next function, which is #4ln(x)#. Again, we know that this derivative is just #4*1/x#

#=>y'=9*1/(4ln(x))*4*1/x*"..."#

Finally, the innermost term is just #x#, which has derivative 1.

#=>y'=9*1/(4ln(x))*4*1/x*1#

#:.y'=9/(xln(x))#