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JEE Main Math Class 12 Continuity and Differentiability Part 1 Questions
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© examsnet.com
Question : 99
Total: 100
If
y
=
(
x
+
√
1
+
x
2
)
n
,
then
(
1
+
x
2
)
d
2
y
d
x
2
+
x
d
y
d
x
is
[2002]
n
2
y
−
n
2
y
−
y
2
x
2
y
Validate
Solution:
Given that
y
=
(
x
+
√
1
+
x
2
)
n
.......(i)
Differentiating both sides w.r. to
x
d
y
d
x
=
n
(
x
+
√
1
+
x
2
)
n
−
1
(
1
+
1
2
(
1
+
x
2
)
−
1
∕
2
.
2
x
)
d
y
d
x
=
n
(
x
+
√
1
+
x
2
)
n
−
1
(
√
1
+
x
2
+
x
)
√
1
+
x
2
=
n
(
√
1
+
x
2
+
x
)
n
√
1
+
x
2
or
√
1
+
x
2
d
y
d
x
=
n
y
[from (i)]
⇒
√
1
+
x
2
y
1
=
n
y
(
∵
y
1
=
d
y
d
x
)
Squaring both sides,
we get
(
1
+
x
2
)
y
1
2
=
n
2
y
2
Differentiating it w.r. to
x
,
(
1
+
x
2
)
2
y
1
y
2
+
y
1
2
.
2
x
=
n
2
.
2
y
y
1
⇒
(
1
+
x
2
)
y
2
+
x
y
1
=
n
2
y
© examsnet.com
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