Consider the equation, xloge(logex)−x2+y2=4 Differentiate both sides w.r.t. x loge(logex)+x⋅x⋅logex1−2x+2ydxdy=0 loge(logex)+logex1−2x+2ydxdy=0 When x=e,y=4+e2. Put these values in (1), 0+1−2e+24+e2dxdy=0 dxdy=24+e22e−1
If x log _e (log _e x )-x^2+y^2=4(y>0), then d y d x at x=e is equal to :
JEE Main 2019 — Mathematics Calculus
If xloge(logex)−x2+y2=4(y>0), then dxdy at x=e is equal to :
Held on 11 Jan 2019 · Verified 6 Jul 2026.
Options
- A
24+e2(1+2e)
- B
24+e2(2e−1)
- C
4+e2(1+2e)
- D
4+e2e
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