How do you differentiate #e^((ln2x)^2) # using the chain rule?

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Answer 1 · 2016-01-17T10:05:44

Use chain rule 3 times. It's:

#2/x*e^((ln2x)^2)#

#(e^((ln2x)^2))'=e^((ln2x)^2)*((ln2x)^2)'=e^((ln2x)^2)*2(ln2x)'=#

#=e^((ln2x)^2)*2*1/(2x)*(2x)'=e^((ln2x)^2)*2*1/(2x)*2=#

#=2/x*e^((ln2x)^2)#

Answer 2 · 2016-01-17T10:09:08

#y' = (2*ln (2x))/x*e^((ln 2x)^2)#

Let #y=e^((ln 2x)^2)#

Differentiate both sides of the equation with respect to x

#(1/y)*y'=2(ln 2x)*1/(2x)*2#