What is the derivative of #f(x)=e^(x^2lnx)#?
1 Answer
Jan 24, 2017
Explanation:
We will use the following:
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The chain rule.
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The product rule.
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#d/dx e^x = e^x# -
#d/dx x^n = nx^(n-1)# -
#d/dx ln(x) = 1/x#
With those:
#= e^(x^2ln(x))(d/dxx^2ln(x))#
#=e^(x^2ln(x))(x^2(d/dxln(x))+ln(x)(d/dxx^2))#
#=e^(x^2ln(x))(x^2(1/x)+ln(x)(2x))#
#=e^(x^2ln(x))(x+2xln(x))#
#=xe^(x^2ln(x))(2ln(x)+1)#
