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For xgt1, " if " (2x)^(2y)=4e^(2x-2y)", ...

For `xgt1, " if " (2x)^(2y)=4e^(2x-2y)", then " (1+log_e2x)^2(dy)//(dx)` is equal to

A

`(x log_(e) 2x + log_(e) 2)/x `

B

` (x log_(e) 2x - log_(e) 2)/x `

C

` x log_(e) 2 x`

D

` log_(e) 2x`

Text Solution

Verified by Experts

The correct Answer is:
B
Given equation is
`(2x)^(2y) = 4*e^(2x-2y)` ….(i)
On applying `'log_(e)`' both sides, we get
`log_(e)(2x)^(2y) = log_(e)4+log_(e)4+log_(e)e^(2x-2y) `
`2ylog_(e)(2x)= log_(e)(2)^(2)+(2x-2y)`
`[:' log_(e) n^(m) = m log_(e) n and log_(e) e^(f(x))=f(x)]`
`rArr (2log_(e)(2x)+2)y= 2x+2 log_(e) (2)`
`rArr" "y=(x+log_(e) 2)/(1+log_(e)(2x))`
On differentiating 'y' w.r.t. 'x', we get
`(dy)/(dx) = ((1+log_(e)(2x))1-(x+log_(e)2)2/(2x))/((1+log_(e)(2x))^(2))`
`=(1+log_(e)(2x)-1-1/x log_(e)2)/((1+log_(e)(2x))^(2))`
So,`(1+log_(e)(2x))^(2)(dy)/(dx)=((x log_(e)(2x)-log_(e) 2)/x)`
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