Continuity and DifferentiabilityHard
Question
If 
= et where t = sin-1 
, then 
is equal to
= et where t = sin-1 , then is equal to Options
A.
B.
C.
D.
Solution
∵ t = 
ln(x2 + y2)
⇒
ln(x2 + y2) = sin-1 
case-I : When x ≥ 0
⇒ ln(x2 + y2) = 2 tan-1
⇒
(2x + 2yy′) = 
⇒ xy′ - yy′ = x + y
⇒ y′ =
let y = x tan θ ; θ ∈
⇒ sin-1
= sin-1
= sin-1
case-II : When x < 0
ln(x2 + y2) = - 2 tan-1
⇒
y′ (x + y) = y - x
y′ =
ln(x2 + y2)⇒
ln(x2 + y2) = sin-1 case-I : When x ≥ 0
⇒ ln(x2 + y2) = 2 tan-1
⇒
(2x + 2yy′) = ⇒ xy′ - yy′ = x + y
⇒ y′ =
let y = x tan θ ; θ ∈
⇒ sin-1
= sin-1= sin-1
case-II : When x < 0
ln(x2 + y2) = - 2 tan-1
⇒
y′ (x + y) = y - x
y′ =
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