Continuity and DifferentiabilityHard
Question
The functions u = ex sin x; v = ex cos x satisfy the equation
Options
A.v 
= u2 + v2
= u2 + v2 B.
= 2 v
= 2 vC.
= - 2 u
= - 2 uD.
= 2v
= 2vSolution
u = ex sin x, v = ex cos x
v
= v(ex cos x + ex sin x) - u(ex cos x - ex sin x)
= ex sin x(v + u) + ex cos x( v - u)
= u(v + u) + v(v - u)
= v2 + u2
again
= ex sin x + ex cos x
= ex sin x + ex cos x + ex cos x - ex sin x
= 2v
similarly other options can be checked.
v
= v(ex cos x + ex sin x) - u(ex cos x - ex sin x)= ex sin x(v + u) + ex cos x( v - u)
= u(v + u) + v(v - u)
= v2 + u2
again
= ex sin x + ex cos x = ex sin x + ex cos x + ex cos x - ex sin x = 2v similarly other options can be checked.
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