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Mathematics
Mathematics
Problem
For
x
>
1
x > 1
x
>
1
, if
(
2
x
)
2
y
=
4
e
2
x
−
2
y
(2x)^{2y} = 4e^{2x-2y}
(
2
x
)
2
y
=
4
e
2
x
−
2
y
, then
(
1
+
log
e
2
x
)
2
d
y
d
x
(1 + \log_e 2x)^2 \frac{dy}{dx}
(
1
+
lo
g
e
2
x
)
2
d
x
d
y
is equal to : [JEE-MAIN ONLINE-2019] (1)
log
e
2
x
\log_e 2x
lo
g
e
2
x
(2)
x
log
e
2
x
+
log
e
2
x
\frac{x \log_e 2x + \log_e 2}{x}
x
x
l
o
g
e
2
x
+
l
o
g
e
2
(3)
x
log
e
2
x
x \log_e 2x
x
lo
g
e
2
x
(4)
x
log
e
2
x
−
log
e
2
x
\frac{x \log_e 2x - \log_e 2}{x}
x
x
l
o
g
e
2
x
−
l
o
g
e
2
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