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BITSAT 2023 Shift 2 Solved Paper
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Section:
Mathematics
© examsiri.com
Question :
111 of 130
Marks:
+1
,
-0
If
y
x
=
e
y
−
x
, then
‌
d
y
d
x
is equal to
‌
1
+
log
‌
y
y
‌
log
‌
y
‌
(
1
+
log
‌
y
)
2
y
‌
log
‌
y
‌
1
+
log
‌
y
(
log
‌
y
)
2
‌
(
1
+
log
‌
y
)
2
log
‌
y
Validate
Solution:
Here,
y
x
=
e
y
−
x
Taking
log
on both sides, we get
‌
log
‌
y
x
=
log
‌
e
y
−
x
‌
‌
‌
(
∵
log
‌
a
b
=
b
‌
log
‌
a
‌
and
‌
log
‌
e
=
1
)
⇒
x
‌
log
‌
y
=
(
y
−
x
)
‌
log
‌
e
⇒
x
‌
log
‌
y
=
y
−
x
On differentiating w.r.t.
x
, we get
‌
d
d
x
(
x
‌
log
‌
y
)
=
‌
d
d
x
(
y
−
x
)
(using product rule)
‌
⇒
x
(
‌
1
y
)
‌
d
y
d
x
+
log
‌
y
(
1
)
=
‌
d
y
d
x
−
1
‌
⇒
‌
d
y
d
x
(
‌
x
y
−
1
)
=
−
1
−
log
‌
y
‌
⇒
‌
d
y
d
x
[
‌
y
(
1
+
log
‌
y
)
y
−
1
]
=
−
(
1
+
log
‌
y
)
‌
⇒
‌
d
y
d
x
[
‌
1
−
1
−
log
‌
y
1
+
log
‌
y
]
=
−
(
1
+
log
‌
y
)
‌
⇒
‌
d
y
d
x
=
−
‌
(
1
+
log
‌
y
)
2
−
log
‌
y
⇒
‌
d
y
d
x
=
‌
(
1
+
log
‌
y
)
2
log
‌
y
© examsiri.com
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