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Joint Entrance Examination
Mathematics
Differentiation
Question 56
Question 56 of 67
Medium
MCQ
If
y
(
x
)
=
x
x
,
x
>
0
y(x)=x^{x},x > 0
y
(
x
)
=
x
x
,
x
>
0
, then
y
′
′
(
2
)
−
2
y
′
(
2
)
y''(2)-2y'(2)
y
′′
(
2
)
−
2
y
′
(
2
)
is equal to
A
\[4(\log_{e}2)^{2}+2\]
B
\[8\log_{e}2-2\]
C
\[4\log_{e}2+2\]
D
\[4(\log_{e}2)^{2}-2\]
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Official solution
y
=
x
x
y
′
=
x
x
(
1
+
ln
x
)
y
′
′
=
x
x
(
1
+
ln
x
)
2
+
x
x
x
f
′
′
(
2
)
−
2
f
′
(
2
)
=
(
4
(
1
+
ln
2
)
2
+
2
)
−
(
2
)
(
4
(
1
+
ln
2
)
)
=
4
(
1
+
(
ln
2
)
2
)
+
2
−
8
=
4
(
ln
2
)
2
−
2
\begin{aligned} & y=x^x \\\\ & y^{\prime}=x^x(1+\ln x) \\\\ & y^{\prime \prime}=x^x(1+\ln x)^2+\frac{x^x}{x} \\\\ & f^{\prime \prime}(2)-2 f^{\prime}(2)=\left(4(1+\ln 2)^2+2\right)-(2)(4(1+\ln 2)) \\\\ & =4\left(1+(\ln 2)^2\right)+2-8 \\\\ & =4(\ln 2)^2-2 \\\\ & \end{aligned}
y
=
x
x
y
′
=
x
x
(
1
+
ln
x
)
y
′′
=
x
x
(
1
+
ln
x
)
2
+
x
x
x
f
′′
(
2
)
−
2
f
′
(
2
)
=
(
4
(
1
+
ln
2
)
2
+
2
)
−
(
2
)
(
4
(
1
+
ln
2
))
=
4
(
1
+
(
ln
2
)
2
)
+
2
−
8
=
4
(
ln
2
)
2
−
2
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