Let `y` be an implicit function of `x` defined by `x^(2x)-2x^xcot y-1=0.` Then `y '(1)` equals:

Options

A. -1
B. 1
C. log 2
D. - log 2

Answer

A

Explanation

` x^(2x) - 2x^(X) cot y - 1 = 0 `
At ` x = 1 " " 1 - 2 cot y - 1 = 0 `
` rArr cot y = 0 rArr y = (pi)/(2)`
On differenting both sides of E.q (i), w.r.t.x, we get
`2x^(2x) (1 + log x) -2[ x^(x) (- "cosec"^(2) y)(dy)/(dx)+ cot y , x^(x) (1 + log x)] = 0 `
` because "If " 4 = x^(2x) , " then log 4" = 2 x log x `
`rArr (1 du)/(4dx) = 2 [ 1 + log x] rArr (du)/(dx) = 2x^(2x) (1 + log x)`
Similarly , ` (d)/(dx) (x^(x)) = x^(x) ( + log x)`
At `(1 , (pi)/(2)) , " "2 (1 + log1) - 2(1(-1)(dy)/(dx))_(((1,(pi)/(2))))+ 0 )=0 `
`rArr 2 + 2((dy)/(dx))_((1","(pi)/(2)))=0`
`rArr ((dy)/(dx))_((1","(pi)/(2)))=-1`