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How do you differentiate # f(x)=xe^((lnx-2)^2 # using the chain rule.?

1 Answer

Nov 1, 2016

#f'(x)=2(lnx-2)e^((lnx-2)^2)+e^((lnx-2)^2)#

Explanation:

#f(x)=xe^((lnx-2)^2)#

Use product rule and chain rule

#f=x, g=e^((lnx-2)^2)#

#f'=1, g'=e^((lnx-2)^2)*2(lnx-2) 1/x#

#f'(x)=fg'+gf'#

#f'(x)=2(lnx-2)e^((lnx-2)^2)+e^((lnx-2)^2)#

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