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

1 Answer
Jothi S.
Dec 5, 2015

#((2sec^2(e^((ln(x)-2)^2) )e^((ln(x)-2)^2)(lnx-2))/x)#

Explanation:

#d/dx (tan(e^((ln(x)-2)^2)))=sec^2(e^((ln(x)-2)^2))*d/dx((e^((ln(x)-2)^2))#
=#sec^2(e^((ln(x)-2)^2))e^(((ln(x)-2))^2 )*d/dx(ln(x)-2)^2#

=#sec^2(e^((ln(x)-2)^2))e^(((ln(x)-2))^2)2(lnx-2)* d/dx(lnx-2)#
=#(sec^2(e^((ln(x)-2)^2))e^(((ln(x)-2))^2)2(lnx-2) *1/x)#
=#((2sec^2(e^((ln(x)-2)^2) )e^((ln(x)-2)^2)(lnx-2))/x)#

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