How do you find the derivative of #ln(x^2)#?

2 Answers
Eddie
Jun 21, 2016

#\frac{2}{x}#

Explanation:

Using the chain rule and letting #u(x) = x^2# we have #\frac{d}{dx} \ln ( u(x)) = \frac{d}{du} \ln ( u(x)) . \frac{du}{dx} = \frac{1}{u(x) }2x#

#= \frac{2x}{x^2} = \frac{2}{x}#

mason m
Jun 21, 2016

#2/x#

Explanation:

There are two methods:

Using the Chain Rule:

Since the derivative of #ln(x)# is #1/x#, we see that the derivative of a function inside the natural logarithm, such as #ln(f(x))#, is #1/f(x)*f'(x)#.

So, for #ln(x^2)#, the derivative is #1/x^2*2x#, since #2x# is the derivative of #x^2#.

Then, we see that #1/x^2*2x# simplifies to #2/x#.

Simplifying first:

Using the rule #log(a^b)=b*log(a)#, we see that #ln(x^2)=2*ln(x)#.

The derivative of #2ln(x)# is just #2# times the derivative of #ln(x)#, which is #1/x#.

We see that #2*1/x=2/x#, the answer we obtained earlier.

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