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If xloge(logex)-x^2+y^2=4(ygt0)",then" d...

If `xlog_e(log_ex)-x^2+y^2=4(ygt0)",then" dy//dx" at " x=e` is equal to

A

` e/sqrt(4+e^(2))`

B

` ((2e-1))/(2sqrt(4+e^(2))`

C

`((1+2e))/(sqrt(4+e^(2))`

D

` ((1+2e))/(2sqrt(4+e^(2))`

Text Solution

Verified by Experts

The correct Answer is:
B
We have, ` x log_(e) (log_(e)x) - x^(2) + y^(2) = 4`, which can be written as
`y^(2) = 4 + x^(2) - x log_(e) (log_(e) x)` …(i)
Now, differentiating Eq. (i) w.r.t. x, we get
` 2y(dy)/(dx) = 2x - x1/(log_(e) x)*1/x - 1* log_(e) (log_(e) x) `
[by using product of derivative]
`rArr" "((dy)/(dx)) = (2x-1/(log_(e)x) -log_(e) (log_(e) x))/(2y) ` ....(ii)
Now, at ` x = e, y^(2) = 4 + e^(2) - e log_(e) (log_(e) e) `
[using Eq. (i)]
`= 4+e^(2) - e log_(e) (1) = 4 + e^(2) - 0`
` = e^(2) + 4 `
` rArr" " y = sqrt(e^(2)+4)" "[:' y gt 0]`
`:. " At " x = e and y = sqrt(e^(2)+4)`,
`(dy)/(dx) = (2e-1-0)/(2sqrt(e^(2)+4))=(2e-1)/(2sqrt(e^(2)+4))` [using Eq. (ii)]
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