Explanation: Let ‌‌y=(‌

1

x

)x
Then, ‌‌log‌y=x‌log(‌

1

x

)=−x‌log‌x
Differentiating both sides w.r.t. x
∴‌‌‌

1

y

‌

dy

dx

‌=−[x⋅‌

1

x

+log‌x]
‌=−(1+log‌x)‌‌‌⋅⋅⋅⋅⋅⋅⋅(i)
On differentiating again eq. (ii), we get
‌

1

y

‌

d2y

dx2

−‌

1

y2

(‌

dy

dx

)2=‌

−1

x

‌‌‌⋅⋅⋅⋅⋅⋅⋅(ii)
From eq. (ii), we get
‌

dy

dx

‌=−y(1+log‌x)‌‌‌⋅⋅⋅⋅⋅⋅⋅(iii)
‌=−(‌

1

x

)x(1+log‌x)
For maximum or minimum values of y, put ‌

dy

dx

=0
Therefore, (‌

1

x

)x(1+log‌x)=0
However, (‌

1

x

)x≠0 for any value of x. Therefore
1+log‌x‌=0
⇒‌‌log‌x‌=−1⇒x=e−1⇒x=‌

1

e

When x=‌

1

e

, from eq. (iii)
‌

1

y

‌

d2y

dx2

−0‌=−e
⇒‌‌‌

d2y

dx2

‌=−e(e)1∕e<0
Hence, y is maximum when x=‌

1

e

and maximum value of y=e1∕e