f′(x)=0 for maximum value
Let y=(x2)x2
lny=x2lnx2
y1y′=2xlnx2+x2x21×x2−2
y′=(xy)(2lnx2−1)
y′=(x2)x2(x)(2lnx2−1)
2lnx2=1
(x2)=e21
x=2e−21
Then maximum value will be
f(2e−21)=(2e−212)4e−1=e2e−1=ee2
The local maximum value of the function, f(x)=(x2)x2,x>0, is
Held on 26 Aug 2021 · Verified 6 Jul 2026.
1
(e4)4e
(e)e2
(2e)e1
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