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Question

for x>1, If (2x)2y=4e2x2y, then
(1+loge2x)2dydx is equal to :

A
xloge2x+loge2x
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B
loge2x
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C
xloge2xloge2x
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D
xloge2x
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Solution

The correct option is C xloge2xloge2x
(2x)2y=4e2x2y
On taking (ln) operator on both sides we get-
2yln(2x)=ln(4)+(2x2y)
y=x+ln21+ln2x
On differentating w.r.t. x
dydx=(1+ln2x)(1+0)(x+ln2)(12x)2(1+ln2x)2
(1+ln2x)2dydx=1+ln2xx+ln2x
(1+ln2x)2dydx=x ln2xln2x

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