JEE Main · 2009 · 12 Apr 2009 · Offline · Differentiation

Let

y

be an implicit function of

x

defined by

x^{2 x} - 2 x^{x} cot y - 1 = 0

. Then

y^{'} (1)

equals

Answer: (A)

Solution

x = 1

the defining relation gives

1 - 2 cot y - 1 = 0

, so

cot y = 0

and

y = \frac{π}{2}

. Differentiating

x^{2 x} - 2 x^{x} cot y - 1 = 0

throughout and then substituting

x = 1

y = \frac{π}{2}

, where

ln 1 = 0

cot \frac{π}{2} = 0

and

csc^{2} \frac{π}{2} = 1

, collapses the expression to

2 + 2 y^{'} = 0

. Hence

y^{'} (1) = - 1

JEE Main · 2009 · 12 Apr 2009 · Offline · Single correct option