for x>1, If (2x)2y=4e2x−2y, then (1+loge2x)2dydx is equal to :
A
xloge2x+loge2x
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B
loge2x
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C
xloge2x−loge2x
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D
xloge2x
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Solution
The correct option is Cxloge2x−loge2x (2x)2y=4e2x−2y
On taking (ln) operator on both sides we get- 2y⋅ln(2x)=ln(4)+(2x−2y) ⇒y=x+ln21+ln2x
On differentating w.r.t. x dydx=(1+ln2x)⋅(1+0)−(x+ln2)(12x)⋅2(1+ln2x)2 ⇒(1+ln2x)2⋅dydx=1+ln2x−x+ln2x ⇒(1+ln2x)2⋅dydx=xln2x−ln2x