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How do you find the derivative of #y=(ln2x)^2#?

1 Answer

Apr 3, 2016

You can do it like this:

Explanation:

#y=(ln2x)^2#

You need to apply the chain rule successively:

#dy/dx=2(ln2x)^(1)xx1/(cancel(2)x)xxcancel(2)#

#dy/dx=(2ln2x)/x#

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